1 


ALGEBRA 

FIRST  COURSE 


0 


IN  MEMORIAM 
FLORIAN  CAJORI 


Digitized  by  the  Internet  Archive 

in  2007  with  funding  from 

IVIicrosoft  Corporation 


http://www.archive.org/details/algebrafirstcourOOIongrich 


CORRELATED    MATHEMATICS    FOR    SECONDARY    SCHOOLS 

ALGEBRA 

FIRST  COURSE 


BT 

EDITH  LONG 

DEPARTMENT  OP  MATHEMATICS,   HIGH   SCHOOL,   LINCOLN,   NEBRASKA 

AND 

W.  C.  BRENKE 

PROFESSOR  OF  MATHEMATICS  IN  THE  UNIVERSITY  OF  NEBRASKA 


NEW   YORK 

THE   CENTURY   CO. 

1913 


Copyright,  1913,  by 
THE  CENTURY  CO. 


TABLE  OF  CONTENTS 

CHAPTER  I 

Paoxs 
Introduction 1-13 

(The  numbera  refer  to  articles.) 

1.  The  Operations  of  Arithmetic.  2.  The  Whole  Numbers,  or 
integers.  3.  Addition.  4.  Subtraction.  5.  MultipUcation. 
6.  Division.  7.  Subject  Matter  of  Algebra.  8.  Algebraic  Ex- 
pressions.    Their  Forms  and  Values. 

CHAPTER  II 

Measurements  —  Lengths,  Angles,  Areas,  Volumes 14r-27 

9.  English  Units  of  Length.  10.  Metric  Units  of  Length. 
11.  Comparison  of  Enghsh  and  Metric  Units  of  Length.  12. 
Angles.  —  Definitions  and  Notations.  13.  Classification  of  An- 
gles. 14.  Measurement  of  Angles.  15.  The  Number  tt.  16. 
Measurement  of  Areas.  17.  Measurement  of  Volumes.  Sum- 
mary. 

CHAPTER  III 

Measurements    Continued.    Temperature,    Weight    and 

Density,  Force 28-41 

18.  Measm-ement  of  Temperature.  19.  English  Units  of 
Weight.  20.  Metric  Units  of  Weight.  21.  The  Beam  Balance. 
22.  The  Lever.  23.  Density  or  Specific  Gravity.  24.  Sum- 
mary.    25.  Questions  and  Problems  for  Review. 

CHAPTER  IV 
Graphic  Representation  of  Quantity 42-48 

26.  The  Amount  of  any  Quantity  Represented  by  the  Length 
of  a  Line. 

CHAPTER  V 
Positive  and  Negative  Numbers 49-54 

27.  Exercises.  28.  Opposite  QuaUties.  29.  Positive  and 
Negative  Numbers.  30.  Graphic  Representation  of  Positive 
and  Negative  Numbers.     31.  Summary, 

ill 


IV  TABLE  OF  CONTENTS 

CHAPTER  VI 

Pagbs 
Addition  and  Subtraction 55-70 

32.  Notation.  33.  Addition  of  Directed  Line  Segments.  34. 
Geometric  Addition  of  Numbers.  35.  Exercises  and  Problems. 
36.  Subtraction.     37.  Summary. 

CHAPTER  VII 

Multiplication  and  Division 71-80 

38.  Meaning  of  Multiplication.  39.  Geometric  Illustra- 
tions of  Multiplication.  40.  Multiplication  of  General  Num- 
bers.    4L  Division.     42.  Summary. 

CHAPTER  VIII 

Addition  and  Subtraction  of  Polynomials 81-89 

43.  Definitions.  44.  Addition  and  Subtraction  of  Poly- 
nomials. 45.  Rules  for  Adding  and  Subtracting  Polynomials. 
46.  Removal  of  Parentheses.     47.  Summary. 

CHAPTER  IX 

Multiplication  and  Division  of  Polynomials 90-104 

48.  Multiplication  of  a  Polynomial  by  a  Monomial.  49.  Dis- 
tributive Law.  50.  The  Commutative  Law  of  Addition.  51. 
The  Commutative  Law  of  Multiplication.  52.  Exponents.  53. 
Multiplication  of  One  Polynomial  by  Another.  54.  Division 
of  Polynomials.     55.  Summary. 

CHAPTER  X 

Problems  Leading  to  Simple  Equations 105-126 

56.  Problems.  57.  Formation  of  an  Equation.  58.  Solu- 
tion of  the  Equation.     59.  Summary, 

CHAPTER  XI 
Simple  Areas  and  Their  Algebraic  Expressions.  Elements 

OF  Factoring 127-149 

60.  Rectangles.  61.  Geometric  Theorems  on  Factoring.  62. 
Parallelograms.  63.  Theorem.  64.  Triangles.  Theorem.  65. 
Regular  Polygons.  66.  Theorem.  67.  Trapezoids.  Theorem. 
68.  Circles.  69.  Factoring  Continued.  70.  Summary.  Exer- 
cises. 


TABLE  OF  CONTENTS  V 

CHAPTER  XII 

Pages 

Fractions 150-165 

71.  Reduction  of  Fractions  to  Lowest  Terms.  Cancellation. 
72.  Multiplication  of  Fractions.  73.  Division  of  Fractions. 
74.  Reduction  to  a  Common  Denominator.  75.  Addition  and 
Subtraction  of  Fractions.  76.  Complex  Fractions.  77.  Long 
Division.     78.  Equation  Involving  Fractions. 

CHAPTER  XIII 

Quadratic  Equations 166-205 

79.  Definitions.  80.  Some  Problems  Leading  to  Quadratic 
Equations.  81.  Square  Roots  of  Numbers.  Pythagorean 
Theorem.  82.  Geometric  Construction  of  Square  Roots.  83. 
Rational  and  Irrational  Numbers.  84.  Factoring  of  Quadratic 
Expressions.  85.  Factors  of  ax^  -\- bx  -\-  c.  86.  Algebraic  So- 
lution of  the  General  Quadratic  Equation,  87.  Imaginary 
Roots.  88.  To  find  the  Square  Root  of  a  Number.  89.  Sum- 
mary and  Problems. 

CHAPTER  XIV 

Variables.    Constants.    Functions  of  Variables.    Graphic 

Representation 206-224 

90.  Variables.     91.  Constants.     92.  Functions  of  Variables. 

93.  Functions  of  a  Single  Variable.    94.  The  Function  Notation. 

95.  Graphic  Representation.     Linear  Functions.    96.  Graph  of 

of  Quadratic  Functions.     97.  Graphs  of  Other  Functions.  98. 

Summary. 

CHAPTER  XV 

Uses  of  the  Graph 225-240 

99.  Graphic  Representation  of  Measurements  of  a  Variable 

Quantity.     100.  Graphic  Solution  of  Equations.     101.  Nature 

of  the  Roots  of  a  Quadratic  Equation.     102.  Graphic  Solution  of 

Equations  of  Higher  Degree.     103.  Summary. 

CHAPTER  XVI 

Loci.     Simultaneous  Equations 241-262 

104.  Meaning  of  the  Word  Locus.  105.  Coordinates.  106. 
Straight  Line  Loci  in  General.  107.  Simultaneous  Equations. 
106.  Summary  and  Problems. 


Vi  TABLE  OF  CONTENTS 

CHAPTER  XVII 

Pages 

Exponents  and  Radicals 263-281 

109.  Definitions.  110.  To  Multiply  a"*  by  a".  111.  To  Mul- 
tiply aT"  by  6"».  112.  To  Divide  oT'  by  a^.  113.  The  meaning 
of  a".  114.  The  Meaning  of  (a*")".  115.  Summary  of  Results. 
116.  Fractional  Exponents.  117.  Irrational  Numbers  or 
Surds.  118.  Examples  Involving  Surds.  119.  Equations  with 
Irrational  Terms.     120.  Summary  and  Exercises  for  Review. 

CHAPTER  XVIII 

Binomial  Theorem 282-283 

121.  The  Binomial  Theorem. 

Protractor Inside  of  back  cover 


PREFACE 

In  the  preparation  of  this  text  the  authors  have  had  in 
view  two  primary  aims,  first  to  emphasize  and  vivify  the 
treatment  of  Algebra  by  a  systematic  correlation  with 
Geometry,  and  secondly  to  present  the  subject-matter  in  a 
style  sufficiently  simple  to  be  easily  grasped  by  students  of 
high  school  age. 

The  first  object  has  been  accomplished  by  a  free  intro- 
duction of  constructive  exercises  from  Geometry,  including 
theorems  and  problems  of  sufficient  range  to  give  to  the 
student  a  fair  working  knowledge  of  the  elementary  prop- 
erties of  important  geometric  figures.  No  attempt  has 
here  been  made  at  formal  demonstration,  that  being  re- 
served for  the  book  on  Geometry  proper;  the  sole  aim  has 
been  to  bring  out  the  fundamental  facts  regarding  such 
figures  as  angles,  intersecting  lines,  polygons,  circles,  and 
some  simple  solids.  These  facts  are  brought  out  by  drawing, 
paper  cutting,  and  super-position,  and  they  are  then  made 
the  basis  of  further  algebraic  work. 

In  bringing  out  the  laws  which  govern  the  four  fundamental 
operations,  systematic  use  is  made  of  graphic  representation 
of  these  operations  on  a  number  scale.  Experience  has 
shown  that  this  is  a  simple  and  effective  way  to  fix  these 
laws  so  firmly  in  the  mind  of  the  student  that  they  are  not 
easily  forgotten. 

A  lengthy  discussion  of  the  contents  of  the  book  would 
be  superfluous  here.  Its  scope,  aims,  and  methods  are 
indicated  by  the  table  of  contents,  and  can  be  more  fully 
appreciated  only  by  reading  the  pages  of  the  text. 

The  second  object,  namely  to  produce  a  simple  treat- 
ment, is  accomplished  largely  by  the  use  of  a  more  nar- 

vii 


Vlll  PREFACE 

rative  style  than  is  usual  in  mathematical  text-books. 
Without  this  the  book  could  have  been  made  considerably 
smaller,  but  it  was  felt  that  the  addition  to  the  number 
of  pages  would  be  more  than  compensated  by  the  gain  in 
clearness  and  fullness  of  explanation.  The  time  and  effort 
required  to  cover  the  year's  work  will  be  less  than  it  would 
be  with  a  more  compact  and  concentrated  form  of  pres- 
entation. 

The  exercises  and  problems  are  drawn  largely  from  the 
student's  own  experience,  and  include  many  that  are  in 
the  nature  of  experiments,  to  be  performed  by  the  student 
individually,  or  with  the  cooperation  of  the  class.  It  is 
urged  that  most  of  these  experiments  be  actually  performed, 
because  of  the  discussion  and  interest  which  comes  from 
such  concrete  applications. 

In  using  the  book  for  the  first  time  there  will  be  perhaps 
some  tendency  on  the  part  of  the  teacher  to  omit  features 
which  at  first  sight  may  seem  unusual.  Such  omissions 
might  easily  detract  materially  from  the  force  and  usefulness 
of  the  text,  and  it  is  strongly  urged  that  they  be  reduced 
to  a  minimum. 

For  those  who  wish  to  make  a  still  closer  correlation 
with  Geometry,  references  to  that  subject  (Part  II  of  this 
series)  are  given,  which  outline  in  a  general  way  a  two 
years'  course  in  Mathematics.  For  this  purpose,  if  the 
teacher  desires  it,  the  two  parts.  Algebra  and  Geometry, 
may  be  had  bound  in  one  volume.  This  will  be  especially 
advantageous  in  schools  where  free  text-books  are  fur- 
nished, because  the  work  of  the  first  year  is  then  always 
at  hand  for  reference  during  the  second  year. 

E.  Long, 

W.  C.  Brenke. 

Lincoln,  Nebraska, 
July,  1913. 


ALGEBRA 

FIRST  COURSE 


CHAPTER  I 
INTRODUCTION 

1.  The  Operations  of  Arithmetic.  In  arithmetic  we  deal 
with  the  nmnbers  1,  2,  3,  .  .  .,  and  with  combinations  of 
these  numbers  by  use  of  the  four  fundamental  operations, 
namely,  addition,  subtraction,  multiplication,  and  division. 
Let  us  briefly  review  the  meaning  of  these  processes. 

2.  The  Whole  Numbers,  or  Integers.  We  first  recall 
that  the  numbers  1,  2,  3,  .  .  .,  10,  11,  12,  called  whole 
numbers  or  integers,  are  merely  used  as  counters,  to  indi- 
cate how  many  objects  are  contained  in  a  given  group. 
The  same  thing  could  be  done  in  many  other  ways,  as,  for 
example,  we  might  make  strokes,  one  for  each  object  in  a 
group.     Thus  we  have  the  Roman  numerals  I,  II,  III,  IV. 

3.  Addition.  If  we  have  two  groups  of  objects,  say  one 
containing  3  and  the  other  4  separate  objects,  their  com- 
bination into  a  single  group  will  contain  how  many  objects? 
Having  counted  the  objects  in  one  group,  say  3,  we  must 
now  count  forward  4  more,  and  so  we  arrive  at  the  number  7. 

We  write  this  result  in  the  form 
3  +  4  =  7. 

This  is  read:  ''Three  plus  four  equals  seven." 

This  means  that  when  three  things  of  any  sort  are  united 
into  a  single  group  with  four  others,  the  result  is  a  group 

1 


2  ALGEBRA  —  FIRST  COURSE 

containing  seven  distinct  objects.  The  process  is  called 
addition. 

The  numbers  3  and  4  are  called  addends,  and  the  number 
7  their  sum. 

Evidently  we  might  reverse  the  order  of  our  counting, 
and  count  first  the  four  objects  and  then  the  three  more  to 
complete  the  group.     This  gives  us 

4  +  3  =  7. 

That  is,  two  numbers  may  be  added  in  either  order;  the 
result  is  the  same. 

In  like  manner  we  interpret  the  equation 

4  +  3  +  5  =  12 

as  the  union  of  three  groups  into  one.  In  all  cases  the  order 
of  writing  the  numbers  to  be  added  together  is  immaterial; 
the  result  is  the  same.  For  example,  we  may  add  a  column 
of  figures  either  up  or  down;  we  may  first  rearrange  the  num- 
bers in  the  column  and  then  add. 

Now  in  applying  the  operations  of  arithmetic  to  actual 
problems,  we  usually  deal  with  objects  of  a  definite  sort. 
We  say  3  apples,  3  seconds,  3  feet,  3  quarts,  and  so  on. 
For  example,  if  we  put  3  oranges  and  4  apples  into  a  basket, 
the  basket  contains  7  objects,  but  neither  7  oranges  nor 
7  apples.  That  is,  if  the  number  which  results  from  the 
addition  of  several  others  is  to  represent  so  many  objects 
of  a  certain  kind,  then  all  the  numbers  which  are  added 
must  represent  objects  of  the  same  kind. 

4.  Subtraction.  The  expression  7  —  3  (seven  minus 
three)  means  that  number  which  when  added  to  3  will 
give  7.  From  arithmetic  we  know  that  this  is  4,  so  we 
have  7  —  3=4.  Subtraction  means  to  find  the  difference 
of  two  numbers,  and  to  do  this  we  find  what  number  must 
be  added  to  the  subtrahend  to  produce  the  minuend. 

Thus  at  a  store  when  you  buy  goods  whose  value  amounts 


INTRODUCTION  3 

to  $1.37,  and  you  pay  with  a  five-dollar  bill,  the  salesman 
hands  you  the  purchased  article  and  then  counts  on  the 
change.  In  other  words,  in  order  to  subtract  $1.37  from 
$5,  he  determines  the  amount  which  must  be  added  to  $1.37 
to  make  $5. 

5.  Multiplication.  The  number  3  can  be  considered  as 
3X1,  that  is,  as  a  unit  taken  three  times  to  measure  the 
quantity  considered.  Thus  3  hours  is  3  X  1  hour;  3  feet  is 
3X1  foot,  and  so  on. 

To  multiply  3  by  4  means  to  take  3  units  4  times,  giving 
12  units;  that  is,  4  X  3  =  12. 

Instead  of  the  cross,  X,  a  dot  is  often  used  to  indicate 
multiphcation;  thus  4  •  3  =  12. 

From  arithmetic  we  know  that  the  product  is  the  same 
when  the  order  of  the  numbers  to  be  multiplied  together  is 
reversed;  that  is,  3  •  4  =  12. 

6.  Division.  To  divide  one  number  by  another  means 
to  find  a  third  number  which  when  multiplied  by  the  divisor 
will  give  the  dividend. 

Thus  12  -T-  3  means  find  a  number  which  when  multi- 
plied by  3  will  give  12.  Our  knowledge  of  multiplica- 
tion enables  us  to  guess  the  right  number  in  many  cases. 
When  we  cannot  easily  guess  the  number,  certain  rules  are 
used  for  getting  it.  We  shall  later  see  the  reason  for  these 
rules,  when  we  study  the  subject  of  division  further. 

The  common  symbols  for  division  are 

io  _!.  o.      12 
12  .  6,       3. 

7.  Subject  Matter  of  Algebra.  In  the  subject  of  algebra, 
we  continue  the  study  of  these  four  fundamental  operations 
as  applied  to  nmnbers,  and  take  up  other  operations.  Also 
we  freely  use  letters  to  designate  numbers  and  it  is  chiefly 
in  its  free  use  of  symbols  that  algebra  differs  from  arith- 
metic. 


4  ALGEBRA  —  FIRST   COURSE 

This  can  best  be  brought  out  by  some  examples. 

Example  1  (a).  If  a  train  travels  5  hours  at  the  rate  of  30  miles  an 
hour,  what  is  the  distance  passed  over? 

Solution:  5  •  30  miles  =  150  miles. 

(b)  If  a  train  travels  t  hours  at  the  rate  of  v  miles  an  hour,  what  is 
the  distance  passed  over? 

Solution:  t  •  v  miles  =  d  miles, 

where  t  stands  for  the  number  of  hours  (time),  v  for  the  rate  per  hour 
(velocity),  and  d  for  the  total  distance. 

Under  (b)  we  have  the  solution  of  every  possible  problem 
of  the  sort  in  (a). 

The  result  is  usually  written  thus 

tv  =  d, 

the  multiplication  sign  being  omitted. 

In  algebra  it  is  understood  that  when  two  letters  are 
written  side  by  side  without  any  sign  between  them,  the 
numbers  for  which  these  letters  stand  are  to  be  multiplied 
together. 

Thus:  a*h  '  c  =  ahc. 

Example  2  (o;.  If  a  rectangle  is  3  feet  wide  and  2  feet  high,  what  is 
its  area? 

Soluiion:  3*2  =  6,  number  of  square  feet  in  the  area. 

(6)  If  a  rectangle  is  b  feet  long  and  h  feet  high,  what  is  its  area? 
•  Solviion:  b  •  h  =  a,  number  of  square  feet  in  the  area. 

Here  again  the  equation  under  (6)  contains  all  possible 
equations  like  those  under  (a). 

Example  3  (a).  A  rectangular  tank,  whose  base  is  6  feet  by  4  feet, 
and  which  is  3  feet  high,  is  filled  with  water  weighing  62^  pounds  per 
cubic  foot.     What  weight  of  water  is  contained  in  the  tank? 

Soluiion:        6  •  4  •  3  =  72,  number  of  cubic  feet  in  the  contents  of 

the  tank. 
72  •  62i  =  4500,  number  of  pounds  in  the  tank. 

(6)  A  rectangular  tank  whose  base  is  a  rectangle  a  feet  by  b  feet, 


INTRODUCTION  5 

and  whose  height  is  h  feet,  is  filled  with  a  Uquid  which  weighs  w  pounds 

per  cubic  foot.    What  is  the  weight  of  the  Uquid  in  the  tank? 

Solution:        a-b'h  =  abh,  number  of  cubic  feet  in  the  contents  of 

the  tank. 

ahh  '  w  =  abhw,  number  of  pounds  in  the  tank. 

If  we  let  W  stand  for  the  number  of  pounds  in  the  tank,  we  have 

the^ormula 

W  =  abhw. 

Note  that  if  w  stands  for  the  number  of  pounds  per  cubic  inch,  abh 
must  be  reduced  to  cubic  inches. 

Example  4  (a).     What  is  the  amount  of  a  principal  of  300  dollars 
placed  at  simple  interest  for  6  years  at  5%? 

Solviion:       j^^  =  number  of  dollars  interest  on  $1  in  1  year. 

300  •  if  7  =    15  =  number  of  dollars  interest  on  $300  in  1  year. 
6  •  15  =    90  =  number  of  dollars  interest  on  $300  in  6  years. 

300  +  90  =  390  =  number  of  dollars,  amount  of  $300  in  6  years. 

(6)  What  is  the  amount  of  p  dollars  placed  at  simple  interest  for  n 
years  at  r%?  . 

Solution:  ^7^  —  number  of  dollars  interest  on  one  dollar  in  one  year. 

p  •  — —  =  YKF.  =  number  of  dollars  interest  on  p  dollars  in  one  year. 
1 UU       J.  uu 

n  •  :^  =  7^  =  number  of  dollars  interest  on  p  dollars  in  n  years. 
100       100 

p  +  -^  =  number  of  dollars  in  the  amount  of  p  dollars  in  n  years. 

If  we  let  a  stand  for  the  number  of  dollars  in  the  amount, 

,  npr 

"=P  +  Io6- 

In  this  formula  replace  p  by  300,  n  by  6,  and  r  by  5;  what  do  you 
get  for  the  value  of  a? 

This  result  is  usually  written  thus: 


'^-A^'-m)' 


where  the  right-hand  member  of  the  equation  is  to  be  understood  as 
follows:  Multiply  the  number  standing  before  the  parentheses,  namely, 
p,  into  each  of  the  numbers  in  the  parentheses,  and  add  the  results. 
Thus:  5(4  +  3-|-2)  =  5.4  +  5-3  +  5-2 

=  45; 
a  (6  +  c  +  d)  =  ab  -\-  ac  -[•  ad. 


6  ALGEBRA  —  FIRST  COURSE 

We  may  first  add  together  the  numbers  within  the  parentheses,  and 
then  multiply  their  sum  by  the  number  outside. 
Thus:  5  (4 +  3  + 2)  =5.9 

=  45. 

In  this  way  we  may  look  upon  6  +  c  +  c?  as  a  single  num- 
ber which  is  to  be  multiplied  by  a.     So  we  may  look  upon 

TIT 

1  +  YqT^  as  a  single  number  to  be  multiplied  by  p. 

Example  5.     What  principal  must  be  put  at  simple  interest  at  6% 
so  that  it  shall  amount  to  $480  in  10  years? 
According  to  Example  4  we  always  have 


Then  we  have 


^(1+^) 


P  = 


1+  — 
^100 


But                                  a  =  480,  n  =  10,  r  =  6. 
Therefore,  p  = j^-^ 

^"^"loo" 

Multiplying  numerator  and  denominator  by  100,  we  have 

48,000 
p  = 


100  +  60 
^  48,000 

160 
=  $300. 
We  here  use  the  fact  that 
if  a  =  b'C,     then     r  =  c. 

No  matter  what  numbers  are  used  for  a,  h,  c,  the  second 
equation  is  always  true  if  the  first  is  true.* 

We  of  course  use  this  rule  as  a  check  on  division  in  arith- 
metic. For  example,  dividing  357  by  17  gives  21.  Check 
the  correctness  of  this  answer  by  showing  that  21  times 
17  gives  357.     That  is,  since  357  =  21  •  17,  then  Vt'  =  21. 

*  There  is  only  one  exception  to  this  rule,  namely,  when  b  is  zero. 
Division  becomes  meaningless  when  the  divisor  is  zero. 


INTRODUCTION  7 

8.  Algebraic  Expressions  —  Their  Forms  and  Values.  In 
the  preceding  examples  we  have  given  some  indication  of 
the  way  in  which  letters  are  used  to  stand  for  numbers 
and  have  made  use  of  some  of  the  notation  which  is  common 
in  Algebra.  We  shall  now  give  a  list,  although  not  a  com- 
plete one,  of  the  various  forms  used  in  the  symbolic  arith- 
metic which  we  call  Algebra.  Much  of  this  is  repetition  of 
what  has  been  said  in  the  examples  above. 

In  what  follows  all  the  letters  and  their  combinations  are 
supposed  to  stand  for  ordinary  arithmetic  numbers. 

(1)  Addition. 

a-{-h  means  to  add  the  number  h  to  the  number  a;  we 
read  it  ''a  plus  6"  just  as  in  arithmetic. 

6  +  a  means  to  add  a  to  6;  we  read  it  "6  plus  a.'' 
Then  we  always  have 

a  +  6  =  b  -\-  a. 
That  is,  the  sum  is  the  same  in  whatever  order  we  add. 

Likewise,  a  +  h  -\-  c  means  the  sum  of  a  and  b  and  c. 
We  read  it  "a  plus  b  plus  c." 

The  sum  is  the  same  in  whatever  way  the  letters  may  be 
written.     The  same  is  true  of  any  number  of  letters. 

(2)  The  expression  (a  +  b)  means  that  we  are  consider- 
mg  the  sum  of  the  numbers  a  and  b  and  are  regarding  it  as 
a  single  number. 

For  example,  the  expression  . 

(a  +  6)  +  (c  +  d) 
means  *Hhe  sum  of  a  and  b  plus  the  sum  of  c  and  d.    But 
this  is  exactly  the  same  as  adding  together  the  four  numbers 
a,  6,  c,  d.     That  is 

(a  +  6)  -f-  (c  +  d)  =  a  +  6  +  c  +  cZ. 
In  the  same  way  {a-{-b  -\-  c)  means  that  we  are  to  regard 
the  sum  of  the  numbers  a,  6,  c  as  a  single  number,  and  so 
on.     Then  we  would  have  for  example 

(a-h6  +  c)  +  (d  +  e)=a  +  6  +  c  +  d  +  e. 


8  ALGEBRA  —  FIRST  COURSE 

In  place  of  parentheses,  (  ),  we  often  use  brackets,  [  ],  or 
braces,   {    J ,  or  the  vinculum. 

Thus  (a  +  b),  [a-\-h],  {a-\-hl,  a +  6;  each  means  that  we 
are  to  regard  the  sum  of  the  numbers  a  and  ?>  as  a  single 
number.     These  marks  are  called  signs  of  aggregation. 

Exercises.  By  giving  arithmetic  values  to  the  letters  in 
the  following,  show  that  the  equations  given  are  true. 

1.  (a  +  6)  +  c  =  a  +  6  +  c.     (Replace  a  by  5,  h  by  2, 
c  by  3.) 

2.  a  +  {h  +  c)  =  a  +  h  +  c.     (Replace  a  by  J,  6  by  J, 
c  by  i) 

3.  (a  +  c)  +  6  =  a  +  6  +  c.     (Replace  a  by  6J,  h  by  8J, 
c  by  51) 

4.  c+  (a  +  h)  =  a-\-b  +  c.  (Replace  a  by  3f ,  h  by  5^, 
c  by  If.) 

5.  {a  +  c)  +  {h  -\-  d)  =  a  +  h  +  c  -\-  d. 

6.  (a  +  d)  +  (c  +  h)  =a  +  h-{-c  +  d. 

7.  {a  +  h  +  c)  +  {d  +  e-\-f)=a  +  h  +  c  +  d  +  e+f. 

8.  (a  +  6)  +  (c  +  c^)  +  (e  +/)  =  a  +  6  +  c  +  d  +  e  +/. 

9.  [m  +  n]  +  [A;  +  Z]  =  m  +  n  +  A;  +  Z. 

10.  [s  -\- 1 -^  u]  +  (v  -\-  w)  =  s  -\-t  +  u-\-v  +  w. 

11.  [s  +  V  -\-  t]  -{-  [w  +  u]  =  s  +  t  +  u  +  V  +  w. 

12.  /i  +  A;  +  b  +  5  +  r]+Sa:  +  2/J=/i-hA;  +  p  +  5  +  r  +  a;  +  2/. 

We  may  now  state  our  first  rule. 

Rule  I.     In  forming  a  sum  the  signs  of  aggregation,  (  ), 

[  ],  {    J ,  ,  may  he  omitted. 

(3)  Subtraction. 

a  —  b  means  the  number  which  must  be  added  to  b  to 
give  a. 

b  —  a  means  the  number  which  must  be  added  to  a  to 
give  6,  as  in  arithmetic. 

We  read  these  as  "a  minus  fe"  and  "b  minus  a"  respec- 
tively. 


INTRODUCTION  9 

Likewise:  a  —  h-\-c  means  ''start  with  the  number  a, 
subtract  the  number  6,  and  to  the  result  add  the  number  c." 
Evidently  this  is  the  same  as  a  +  c  —  6. 

The  signs  of  aggregation  are  used  as  before. 

Thus:  (a  +  6)  +  (c  —  d)  means  "to  the  sum  of  a  plus  h 
add  the  difference  of  c  minus  rf."  This  is  the  same  as 
a  -\-b  -r  c  —  d,  that  is  we  may  omit  the  parentheses. 

(a  +  6)  —  (c  —  d)  means  "from  the  sum  of  a  and  h  sub- 
tract the  difference  of  c  minus  c?."  But  we  cannot  remove 
the  parentheses  and  say  our  answer  is  the  same  as  a  +  & 
—  c  —  (i,  as  a  simple  example  will  show. 

We  have  (10  +  7)  -  (12  -  4)  =  17  -  8 

=  9. 
But  10  +  7  -  12  -  4  =  1. 

The  two  results  are  not  the  same,  so  that  in  subtraction  we 
cannot  drop  the  parentheses  as  in  addition.  We  shall  soon 
have  a  rule  to  cover  this  case. 

Exercises.  Remove  as  many  as  possible  of  the  signs  of 
aggregation  in  the  following  exercises. 

1.   (a-h)  +  (c-d)  =  ?  2.   (b-c)  +  id-  a)  =? 

3.  ia-\-b-c)-d=?  4..  d-\-{b-c-d)  =? 

5.  [a-d-{-c]-\-[d-e]  =  ?  6.   [a +  b] -{- [d  -  e  -  c]  =  ? 

7.    \h-kl+[m-n]  =  ?  8.   s -\- t  +  u  -  v -{- p  -  q  =  ? 

9.   (f-\-9)-(h-j)  +  k^'^  10.   (u+v)-(x±y)-(u-z)  =  l 

11.  n-{p  +  q)-{s-t)  =  ?  12.   (a-6)-c  +  d-[e+/]  =  ? 

Show  by  inserting  arithmetic  numbers  for  letters  that 
the  expressions  in  the  last  four  exercises  change  their  values 
when  the  signs  of  aggregation  are  omitted. 

(4)  Multiplication. 

aXb  means  to  multiply  the  number  b  by  the  number  a. 
Since  this  is  the  same  as  multiplying  a  by  b,  we  have 

aXb  =  bXa, 


10  ALGEBRA  —  FIRST  COURSE 

Using  the  dot  to  indicate  multiplication, 

aXb=a'b=b'a. 

Usually  we  do  not  use  either  the  cross  or  the  dot;    we 
simply  write  ah,  thus: 

aXh  =  a'h  =  ah  —  ha. 

Notice  that  this  is  contrary  to  the  rule  in  arithmetic: 

2i  =  2  +  i,     not  2  X  i 

The  numbers  a  and  6  are  called  factors,  and  ah  is  their 
product. 

Similarly  we  have 

aXh  X  c  =  a'h  *  c  =  ahc, 

and  the  "factors"  a,  h,  c  may  be  taken  in  any  order.     Thus: 

ahc  =  ach  =  bca,  and  so  on. 

This  applies  to  the  product  of  any  number  of  factors. 
We  also  may  use  the  signs  of  aggregation: 

aX{b-\-c)  =a{b  +  c) 
=  ah  +  ac. 
aX[h  +  c  -  d]  =  a\h  -\-  c  -  d] 
==  ah  +  ac  —  ad. 

(5)  Division.     The  symbol  j-  means  the  quotient  of  a 

divided  by  h,  this  quotient  being  a  number  which  when 
multiplied  by  h  will  give  a. 

In  place  of  t  we  may  use  a-r-h. 

If  we  call  the  quotient  q,  then  since  the  quotient  multiplied 
by  the  divisor  equals  the  dividend,  we  have, 

a  =  bq  equivalent  to  ^  =  q- 

Let  the  student  carefully  pote  our  definition: 


I,  §  8]  INTRODUCTION  1 1 

TJie  operation  of  dividing  ahy  b  consists  in  finding  a  number 
q  which  if  multiplied  by  b  gives  a.  That  is,  r  or  a-r-b,  each 
stands  for  that  number  q,  sv^h  that  bq  =  a. 

The  expression  r  is  called  a  fraction,  a  being  the  numer- 
ator and  b  the  denominator.     The  numbers  a  and  b  are 
called  the  terms  of  the  fraction. 
Then,  as  in  arithmetic,  we  have 

2  _2j^  =  !LL?. 

3  2-3      n-S' 

so  we  may  have  a  in  place  of  2  and  b  in  place  of  3,  and  may 
write 

a  _3a  _na 

b~Sb~nb' 

Rule.  The  two  terms  of  a  fraction  may  be  multiplied  by 
the  same  number  without  changing  the  value  of  the  fraction. 
That  is, 

.<.  a  .,  na 

if  T  =  Qf     then     -r  =  q. 
b      ^  nb      ^ 

(6)  Cancellation.  If  both  terms  of  a  fraction  contain 
the  same  factor,  this  common  factor  may  be  cancelled  out. 
Thus, 

y^a  _  a 
^b~b' 

(7)  Division  of  Products.  When  a  product  of  numbers 
is  to  be  divided  by  a  number,  or  by  a  product  of  numbers, 
common  factors  should  first  be  removed  by  cancellation. 

144  »  56  -  18  _  ^'^'^'?'?-?'f'?'2'2'^'S'2  _ 
36-12.14  ^.^.^.^.^.^.^.y.^  '^' 


12  ALGEBRA  —  FIRST  COURSE  tt.$8 

Likewise 

24 amnr  _  3  •  ^  •  ^  •  ? amyir 
8  hn  %^%*%h^ 

3amr 


Especial  attention  is  called  to  the  fact  that  if  you  are 
called  upon  to  divide  a  given  number  by  a  certain  number 
and  then  to  multiply  by  that  same  number,  the  result  will 
be  the  given  number.  Thus,  if  you  are  told  to  divide  629 
by  250  and  then  to  multiply  by  250  the  result  is  629.  Or, 
in  arithmetic  form, 

Aoq 
250-250  =  629. 


In  all  cases, 


a*-  =  0, 


no  matter  what  the  values  of  a  and  h. 

The  rule  that  you  can  multiply  the  numerator  and  de- 
nominator by  the  same  number  and  not  change  the  value 
of  the  fraction,  gives  us  the  simplest  method  of  reducing  a 
complex  fraction  to  a  simple  one.  You  have  but  to  find 
the  least  common  denominator  of  all  the  fractions  found 
in  both  numerator  and  denominator,  and  multiply  both 
terms  of  the  fraction  by  it.    Thus, 

3f 
given  ^^• 

The  least  common  denominator  of  the  fractions  is  15. 
Multiply  the  numerator  and  denominator  by  15: 

^^55 
2/ir      37; 

A  little  practice,  to  make  the  student  familiar  with  this 
method,  is  worth  while. 


I.  §  8]  INTRODUCTION  13 

Exercises.     Simplify  the  following  fractions: 

11  3  1  _L  1 

1.     -0-*  ^'     T  ^'      1      I      2 


81|'  f  -  iV     ■  824 


Exercises  for  Review.  Find  the  numerical  value  of  each  of 
the  following  expressions  when  the  letters  are  replaced  by 
the  given  numbers. 


1.   a  +  h  +  c; 

o  =  3,  6  =  4,  c  =  5. 

2.   a  —  h  +  c; 

a  =  10,  6  =  3,  c  =  4. 

3.    (a  +  6)H-(c-c^); 

a  =  21,  6  =  15,  c  =  40,  d  =  28. 

4.    (a-h)-{c  +  d); 

a  =  93,  6  =  22,  c  =  12,  d  =  18. 

6.   p(q  +  r  -  s); 

p  =  15,  g  =  65,  r  =  32,  s  =  12. 

6.  (p  +  g)  (p-  q); 

P  =  n,q  =  M. 

^'  h-k' 

^=5'^=6- 

mn 

Q                                      « 

'   m-\-n' 

3           15 

.    x-\-(y-z), 
^'  x-(y  +  z)' 

3           2           1 

^=2'2/=3''=6- 

10.  ,^^^,     ;  p  =  56,  g  =  112,  r  =  28. 

i+f 

11.   ;  a  =4,  6  =  5. 


12. 
13. 


14. 


'-f 

[r+(s-t)]- 

(s  +  t): 

;  r  =  i,s  =  i,t  =  h 

a  +  (6  -  c) . 
a-(6  +  c)' 

TO      n 
n      m 
mn    ' 

3           2 

CHAPTER  II 

MEASUREMENTS  — LENGTHS,    ANGLES,    AREAS, 
VOLUMES 

9.  English  Units  of  Length.  You  are  familiar  with  the 
yardstick,  the  foot  rule,  and  the  inch.  To  tell  some  one 
how  long  a  certain  line  is  you  might  say  that  it  measures  so 
many  yards;  the  yard  is  then  your  unit  of  length.  To 
state  the  length  of  a  shorter  line  you  might  tell  the  number 
of  feet  which  it  contains;  the  foot  is  then  your  unit  of  length; 
for  a  still  shorter  line,  such  as  you  might  draw  on  a  sheet  of 
paper,  the  inch  would  be  a  convenient  unit.  Quite  long 
lines,  as  the  length  of  fence  around  a  farm,  are  measured 
with  the  rod  as  the  unit  of  length;  for  distances  between 
cities  we  use  the  mile  as  the  unit. 

'    So  we  have  the  following  common  English  units  of  length: 
inch,  foot,  yard,  rod,  mile. 

Exercise.  How  many  feet  in  3  yards?  In  5|  yards? 
In  n  yards?  How  many  inches  in  each  of  these?  How 
many  rods  in  1  mile?     In  2j  miles?     In  m  miles? 

10.  Metric  (or  French)  Units  of  Length.  The  French 
have  chosen  a  different  set  of  units  to  measure  lengths  of 
lines.  This  is  based  on  the  decimal  system  and  is  com- 
monly used  in  scientific  work.  In  this  system  there  is  a 
unit  corresponding  nearly  to  the  English  yard,  and  called  a 
meter.  It  is  a  little  longer  than  the  yard  and  represents 
one  ten-millionth  part  of  the  distance  from  the  earth's 
equator  to  the  pole,  measured  along  a  meridian.  One  tenth 
of  a  meter,  about  four  inches,  is  called  a  decimeter;  one 
hundredth  of  a  meter  is  a  centimeter,  a  little  less  than  half 

14 


II.  §  11] 


MEASUREMENT  OF  LENGTHS 


15 


an  inch;  one  thousandth  of  a  meter  is  a  millimeter.  For 
long  lines,  where  we  would  use  the  mile,  the  French  unit  is 
the  kilometer,  or  one  thousand  meters,  a  little  more  than 
half  a  mile. 

Thus  the  common  metric  units  of  length  are: 
millimeter,  centimeter,  decimeter,  meter,  kilometer. 

Exercise.  How  many  decimeters  in  half  a  meter?  In 
2.3  meters?  In  m  meters?  How  many  centimeters  in  each 
of  these? 

11.  Comparison  of  English  and  Metric  Units  of  Length. 
The  figures  below  show  parts  of  a  ruler,  one  edge  of  which 
is  divided  into  inches  and  tenths  of  inches,  the  other  into 
centimeters  and  millimeters.  Such  a  ruler  will  be  found 
inside  the  back  cover  of  this  book. 


MM     MM 

Inches  and  Tenths 

M  M  1     MM     M  M  M  M  1 

M 

1  1  1 

1 
1                9. 

s           h         '5 

6                7] 

8 

9 

WMMMm 

MMMiniMI 

II  III!  III!  nil  nil 

Hill 

II  Hill 

Centimeters  and  Millimeters 


Inches  and  Tenths 


I  I  I  M  I  M  II  M  M  M  I  M  M  M  I  M  M  M  M  M  I  I 


33  \ 
83  \  8U\ 


851 


36 
91] 


Centimeters  and  Millimeters 


Exercise  1.  From  the  first  figure  read  off,  as  exactly  as 
possible,  the  number  of  inches  in  a  centimeter;  the  number  of 
centimeters  in  an  inch. 

Exercise  2.  Draw  a  line  several  inches  long.  Measure  it 
with  the  edge  of  the  ruler  marked  in  common  units;  measure 
it  again  in  metric  units.  From  these  measurements  find 
the  number  of  inches  there  are  in  a  decimeter.  Solve  as  in 
the  following  example. 


16  ALGEBRA  —  FIRST  COURSE  [ii.§i2 

Example.  Suppose  the  line  measures  4.5  inches  and  also  1.14  deci- 
meters.    To  find  the  number  of  inches  in  one  decimeter: 

Let  i  =  the  number  of  inches  in  one  decimeter. 

Then  1.14  i  =  the  number  of  inches  in  1.14  decimeters. 

But  4.5  =  the  number  of  inches  in  1.14  decimeters; 

therefore,        1.14  i  =  4.5,  since  both  stand  for  the  measurement  of  the 
same  line. 

Therefore  i  =  3.95.     (Why?) 

So  we  have  3.95  inches  in  a  decimeter,  as  nearly  as  we  can 
tell  from  the  given  measurements. 

Repeat  this  experiment,  using  different  lengths  of  lines. 
Exercise  3.     From   the   second  figure  above,  how  many 
inches  in  9  decimeters?     Hence  how  many  inches  in  one 
decimeter? 

Exercise  4.  Draw  a  line  2.75  inches  long.  Measure  it  in 
centimeters.  Now  solve  as  above  to  find  how  many  centi- 
meters in  an  inch.     Let  c  stand  for  the  required  number. 

Exercise  5.  Draw  a  line  0.75  inch  long.  Measure  it  in 
millimeters.  Solve  as  above,  to  find  the  number  of  milli- 
meters in  an  inch.  How  does  this  answer  compare  with  the 
answer  to  Exercise  4? 

Exercise  6.  A  line  measures  x  inches.  How  many  deci- 
meters does  it  contain? 

Exercise  7.  A  line  measures  y  centimeters.  How  many 
feet  does  it  contain? 

12.*  Angles  —  Definitions  and  Nota- 

/^  tion.     Suppose  that  we  partially  open 

,///^  a  fan,  or  two  of  the  arms  of  a  folding 

^/^''  ruler.     Suppose   also   that   a   line   is 

/^^ drawn  on  each  arm  of  the  ruler,  start- 

O-- — -" IJ    ij^g  ixom  the  pivot  and  following  the 

middle  of  the  arm. 
Definitions,    The  figure  so  formed  by  two  straight  lines 
starting  out  from  the  same  point  is  called  an  angle.     The 

*  At  this  point  Chapter  I  of  Geometry  may  be  taken  up  if  closer 
correlation  is  desired. 


II.  §  13] 


MEASUREMENT  OF  ANGLES 


17 


point  marked  A  in  the  figure  (the  pivot)  is  called  the  vertex 
of  the  angle.  The  lines  AB  and  AC  are  called  the  arms  of 
the  angle. 

Notation.  We  shall  often  use  a  single  letter,  usually  a 
capital,  placed  near  a  point  in  a  figure  to  designate  that 
point.  If  we  designate  a  certain  point  by  A,  and  another 
point  by  B,  the  straight  line  through  these  two  points  is 
called  the  line  AB  or  the  line  BA.  We  would  say  'Hhe  line 
AB"  when  the  line  is  drawn  from  A  to  B;  we  would  say 
"the  line  BA  "  when  the  line  is  drawn  from  B  to  A.  Often 
it  makes  no  difference  which  we  use;  in  other  cases  a  dis- 
tinction is  necessary. 

To  designate  the  angle  formed  by  the  lines  AB  and  AC, 
we  say  ''the  angle  BAG/'  or,  ''the  angle  CAB.''  The  first 
means  that  in  opening  up  the  angle  we  regard  AB  as  a  fixed 
arm  and  AC  as  revolving;  the  second  means  that  AC  is 
the  fixed  arm  and  that  AB  is  revolving.  Often  it  makes  no 
difference  which  notation  is  used.  In  either  case,  to  desig- 
nate an  angle,  first  name  a  point  on  the  fixed  arm,  then 
name  the  vertex,  and  finally  name  a  point  on  the  movable 
arm. 

In  fixed  figures,  angles  may  usually  be  read  either  way. 

The  symbol  for  the  word  "angle"  is  Z  ;  so  ABAC  means 
"angle  BAC." 

Exercise.  In  each  of  the  figures  below  read  off  the  points, 
lines,  and  angles  marked  in  it. 


13.   Classification  of  Angles.     Angles  are  classified  accord- 
ing to  the  amount  of  turning  done  in  separating  the  arms. 


18 


ALGEBRA  — FIRST  COURSE 


tll,  §  14 


In  the  figure  on  p.  16  suppose  AB  to  be  fixed  and  AC 
to  be  revolving;  when  AC  has  revolved  half  way  around,  so 
that  it  lies  just  opposite  to  AB  and  forms  one  straight  line 
with  it,  the  angle  BAC  is  called  a  straight  angle;  half  of  a 
straight  angle  is  a  right  angle;  if  AC  turns  through  less  than 
a  right  angle  it  forms  with  AB  an  acute  angle;  if  AC  turns 
through  more  than  a  right  angle  but  less  than  a  straight 
angle,  it  forms  with  AB  an  obtuse  angle ;  more  than  a  straight 
angle  is  called  a  reflex  angle;  a  complete  turn  is  called  a 
perigon. 


Z2\ 


C.  A  -B    A  B     A  JB 

Straight  Angle  BAC.  Right  Angle  BAC.  Acute  Angle  BAC. 


\ 


X\ 


A.  B 

Obtuse  Angle  BAC. 


Perigon  BA  C. 


Reflex  Angle  BAC. 

Exercise.  Classify  each  of  the  angles  in  the  figures  on 
p.  17. 

14.  Measurement  of  Angles.  To  state  the  size  of  an 
angle  we  adopt  some  standard  angle  as  a  unit,  and  say  how 
many  of  these  units  are  needed  to  fill  the  given  angle.  Two 
different  units  are  in  common  use,  one  called  the  degree 
the  other  the  radian. 

Degree  Measure  of  Angles.  When  a  perigon  is  divided 
into  360  equal  parts,  each  such  part  is  called  a  degree. 


II.  §  14]  MEASUREMENT  OF  ANGLES  19 

So  we  have 

360  degrees  =  a  perigon. 
Then  180  degrees  =  a  straight  angle 

and  90  degrees  =  a  right  angle. 

The  symbol  for  degrees  is  °,  so  that  10°  means  "ten 
degrees." 

Angles  are  usually  measured  with  a  protractor  (see  inside 


Protractor 

of  back  cover)  as  shown  in  the  above  figure.  To  measure, 
a  reflex  angle,  measure  its  excess  over  a  straight  angle,  or 
measure  what  is  lacking  to  make  a  perigon. 

Exercise  1.  Draw  ten  different  angles,  some  acute,  some 
obtuse,  and  some  reflex.  Measure  each  and  write  its  value 
on  your  figure. 

Exercise  2.  Draw  a  triangle.  Measure  each  angle. 
What  is  their  simi?  Repeat  this  with  another  triangle  of 
different  shape. 


Exercise    3.     Draw  a  triangle  and  tear  apart  as  in  the 
above  figure.     Place  the  three  angles  with  their  vertices 


20  ALGEBRA  — FIRST  COURSE  [ii,  §  15 

together,  one  angle  next  to  the  other,  without  overlapping. 
What  is  the  sum  of  the  angles  of  the  triangle? 

Exercise  4.  Repeat  Exercise  2,  using  a  quadrilateral, 
that  is,  a  figure  bounded  by  four  straight  lines.  Do  not 
draw  a  square  or  a  rectangle,  but  rather  a  figure  whose 
sides  and  angles  are  quite  unequal.] 

Exercise  5.     Repeat  Exercise  3,  using  a  quadrilateral. 

Exercise  6.  Repeat  Exercises  2  and  3,  using  a  pentagon, 
that  is,  a  figure  bounded  by  five  straight  lines. 

Radian  Measure  of  Angles.  In  this  system  the  unit  of 
measure  is  a  radian,  instead  of  a  degree  as  in  the  system 
just  considered.  You  can  easily  make  a  protractor  gradu- 
ated in  radians. 

Exercise  1.  On  stiff  paper,  or,  better,  light  cardboard, 
draw  a  circle  with  a  radius  of,  say,  two  inches.  Carefully 
cut  it  out  and  mark  a  point  on  the  circumference  or  rim. 
On  a  good-sized  sheet  of  paper  draw  a  straight  line  and  mark 
off  on  it  parts,  each  equal  to  the  radius  of  the  circle.  Roll 
the  circle  carefully  along  this  line,  starting  with  the  marked 
point  on  the  rim  placed  at  the  beginning  of  the  first  division 
on  the  line.  Each  time  that  a  point  on  the  rim  of  the  rolling 
circle  reaches  a  division  point  on  the  line,  mark  that  point 
on  the  rim.  Now  draw  lines  from  the  center  of  the  circle 
to  the  points  marked  on  the  rim.  You  then  have  a  series 
of  equal  angles,  each  of  which  is  one  radian. 

Exercise  2.     Define  a  radian. 

Exercise  3.  By  rolling  the  circle  so  that  it  makes  just 
one  complete  turn,  find  approximately  how  many  radians 
there  are  in  a  perigon.  You  will  find  a  little  more  than 
six  radians.  Estimate  the  decimal  part  as  well  as  you 
can. 

15.  The  Number  it.  .  The  number  of  radian  units  in  a 
perigon  is  not  a  whole  number,  as  you  found  in  the  last  exer- 
cise. Nor  can  this  number  be  expressed  either  by  a  fraction 
or  by  a  terminating  decimal.     It  is  a  so-called  incommen- 


II.  §15]  MEASUREMENT  OF  ANGLES  21 

surable  number  and  by  general   agreement  is  always  indi- 
cated by  2  TT,  TT  being  a  Greek  letter  called  ''pi." 
We  therefore  have 

2  TT  radians  =  a  perigon  =  360  degrees. 
TT  radians  =  a  straight  angle  =  180  degrees. 

180 
Then  1  radian  =  — degrees.     (About  57°.3.) 

TT 

The  number  for  which  w  stands,  to  four  decimal  places,  is 
3.1416;  less  exactly  it  is  ^j^-.  It  should  be  remembered  that 
both  of  these  values  are  only  approximate. 

Exercises. 

1.  From  Exercise  3  above,  find  as  nearly  as  you  can  the 
number  of  radii  that  must  be  taken  to  equal  the  length  of 
the  circumference  of  a  circle.  The  number  of  diameters. 
(A  diameter  of  a  circle  is  a  line  through  the  center  and 
terminated  each  way  by  the  circumference.) 

2.  If  c  denotes  the  circumference  of  a  circle,  d  the  diam- 
eter, and  r  the  radius,  show  that 

c  =  2  TTr, 

and  that 

C  =  TTd, 

State  these  equations  in  words  and  learn  them. 

3.  Using  the  radian  scale  on  your  protractor,  repeat  Exer- 
cise 2  (p.  19). 

4.  Using  the  radian  scale  on  your  protractor,  repeat 
Exercise  4  (p.  20). 

5.  Using  -2y2-  as  the  value  of  tt,  find  the  number  of  degrees 
in  one  radian.  Repeat,  taking  tt  as  3.1416.  In  each  case 
carry  the  result  to  three  decimal  places. 

6.  Draw  an  acute  angle.  Measure  it  with  the  degree 
scale  on  your  protractor,  then  with  the  radian  scale.  Solve 
as  in  the  example  worked  out  on  p.  16  to  find  tKe_ number 
of  degrees  in  a  radian. 


22 


ALGEBRA  —  FIRST  COURSE 


III.  §  16 


7.  Draw  an  obtuse  angle  and  do  the  work  as  instructed 
in  Exercise  6. 

8.  Draw  a  reflex  angle  and  repeat  work  of  Exercise  6. 
How  do  the  answers  to  Exercises  6,  7,  8  compare?    How 

do  they  compare  with  results  of  previous  work? 

9.  How  many  radians  are  there  in  the  sum  of  the  angles 
of  a  pentagon  ? 

16.  Measurement  of  Areas.  If  a  unit  of  length  is  chosen, 
then  a  square  whose  side  is  that  unit  of  length  is  our  unit 
of  area.  To  measure  an  area  we  must  find  how  many  such 
square  units  are  required  to  cover  it. 

Exercise  1.  On  cross-section  paper  draw  a  rectangle  of 
any  convenient  base  and  height.  How  many  linear  units 
are  there  in  the  base?  In  the  height?  How  many  square 
units  in  the  area?  To  answer  the  last  question  count  the 
squares.     Do  your  numbers  fit  the  formula 

Area  of  rectangle  =  base  times  height? 

In  letters,  this  formula  may  be  written 
a  =  bh. 

Verify  this  formula  by  drawing  several  other  rectangles 
such  as:  a  =  3,  6  =  4;  a  =  5,  6  =  2;  a  =  J,  6  =  J. 
When  the  rectangle  is  a  square  whose  side  is  h,  its  area  is 

a  =  hh. 
In  place  of  hh  we  usually  write  b^,  so  that 

a  =  h\     (a  equals  the  square  on  6.) 


Exercise  2.     Draw  a  parallelogram,  that  is,  a  quadrilateral 
whose  opposite  sides  are  parallel.     Cut  this  out  from  your 


II.  §  16]  MEASUREMENT   OF  AREAS  23 

paper.  Now  cut  off  a  triangle  from  one  side  and  fit  it  on 
the  other.  Is  the  area  of  your  parallelogram  equal  to  the 
area  of  a  rectangle  of  the  same  base  and  height? 

Repeat  Exercise  1,  using  a  parallelogram.  So  verify  the 
statement: 

Area  of  parallelogram  =  base  times  height. 
Using  letters:  a  =  bh, 

as  for  the  rectangle. 

Exercise  3.  On  cross-section  paper  draw  a  triangle,  tak- 
ing one  of  its  sides  along  one  of  the  ruled  lines.  Call  this 
side  the  base.  Draw  a  line  at  right  angles  to  the  base  and 
leading  to  the  opposite  comer  of  the  triangle.  Call  this 
line  the  altitude.  How  many  units  are  there  in  the  length 
of  the  base?  In  the  altitude?  Find  as  well  as  you  can 
the  number  of  square  units  in  the  area  of  the  triangle,  add- 
ing parts  of  squares  to  make  whole  squares.  Do  your  num- 
bers fit  the  formula 

Area  of  triangle  =  J  (base  times  height)? 

Draw  several  other  triangles  and  in  each  case  verify  the 
formula 

a  =  ^hh. 

Exercise  4.  Draw  a  circle  on  your  paper;  also  draw  a 
triangle  with  its  base  equal  in  length  to  the  circumference 
of  the  circle  and  its  height  equal  to  the  radius  of  the  circle. 
Count  the  number  of  squares  in  each  and  see  how  they 
compare,  approximately. 

How  would  you  find  the  area  of  a  circle?  Letting  a  stand 
for  the  number  of  square  units  in  the  area,  c  for  the  number 
of  linear  units  in  the  circumference,  and  r  for  the  number  of 
linear  units  in  the  radius,  we  have  the  statement 

a  =  ^cr; 
but  we  have  shown  that  c  =  27rr; 
therefore  a  =  J  •  2  irr  •  r 

=  irr\ 


24 


ALGEBRA  —  FIRST  COURSE 


[11,  §  17 


Exercise  5.  On  cross-section  paper  draw  a  circle  with 
any  convenient  radius,  such  as  five  times  the  side  of  one 
square.  Find  as  nearly  as  possible  the  number  of  square 
units  in  the  area  of  the  circle.  Also  find  the  area  of  a  square 
whose  side  equals  the  radius  of  the  circle.  Do  your  num- 
bers fit  the  formula 

Area  of  circle  =  tt  times  square  on  radius? 

Draw  several  other  circles  of  various  sizes  and  in  each 
case  verify  the  formula 

a  =  Trr^. 

17.  Measurement  of  Volume.  To  measure  the  cubical 
content  of  a  solid  body  we  choose  as  our  unit  of  measure  a 
cube  whose  edge  is  one  linear  unit.  The  number  of  such 
cubic  units  contained  in  the  body  to  be  measured  is  called 
its  volume. 

For  example,  a  pint  jar  will  contain  about  27  cubic  inches 
of  water;  that  is,  its  volume  is  27  cu.  in. 

The  volumes  of  a  few  regular  bodies  are  given  below,  each 
in  terms  of  the  dimensions  of  the  body.  The  dimensions  in 
each  formula  should  be  expressed  in  the  same  unit  of  length. 
State  each  formula  in  words. 

Exercises. 

1.  Measure  the  dimensions  of  a  rectangular  parallelopiped. 
Find  its  volume  by  immersing  it  in  water  in  a  graduated 
beaker  and  noticing  how  much  the  water  rises.  Do  your 
numbers  fit  the  formula? 


/ 


I 

L. 


....i 


Bectanoular  Parallelopiped 
Volume.'^  dhc. 


Cylinder 
Volume  ^Ti'T'^  b. 


II.  §  17] 


MEASUREMENT  OF  VOLUME 


25 


Cone 
Volume^  ys"^^^^' 


Sphere 
Volume  ^ya'^'''^. 


(6)  Cylinder. 

r  =  S,h  = 

=  2; 

r 

=  1 

(c)  Sphere. 

r  =  5; 

r 

=  3§ 

2.  Repeat  Exercise  1  with  a  cylinder. 

3.  Repeat  Exercise  1  with  a  cone. 

4.  Repeat  Exercise  1  with  a  sphere. 

5.  Calculate  the  volume  when  the  dimensions  are  as 
below.     (Take  tt  as  V"  or  3|.) 

(a)  Cone. 

r  =  2,  /i  =  4;    r  =  1|,  /i  =  2J;    r  =  3i  /i  =  f . 

r  =  2f. 

6.  Divide  the  volume  of  a  cylinder  by  the  volume  of  a 
cone  having  the  same  base  and  height.  What  is  the  quo- 
tient? 

7.  If  a  cylindrical  glass  is  filled  with  water  and  then  a 
solid  cone  of  the  same  base  and  height  is  placed  in  the  glass, 
what  part  of  the  water  will  be  forced  out? 

8.  A  cone-shaped  funnel  is  8  inches  across  the  top  and  6 
inches  deep.     How  much  water  will  it  hold? 

9.  Measure  the  dimensions  (height  and  radius)  of  a  pint 
jar,  using  the  inch  as  the  unit  of  length.  How  many  cubic 
inches  of  water  will  the  jar  hold?     Is  "a  pint  a  pound"? 

10.  A  sphere  and  a  cylinder  have  the  same  radius.  How 
high  should  the  cylinder  be  to  have  the  same  volume  as  the 
sphere?  First  take  the  radius,  say,  3  inches;  then  solve 
again  taking  the  radius  equal  to  r. 


26  ALGEBRA  —  FIRST  COURSE  [Ii.§i7 

Summary. 

Measurement  of  Length 

English  Units  *  Metric  Units 

Inch  (in.)  Millimeter    (mm.) 

Foot  (ft.)  Centimeter  (cm.) 

Yard  (yd.)  Decimeter    (dm.) 

Rod  (rd.)  Meter  (m.) 

Mile  (mi.)  Kilometer     (km.) 

1  rod  =  16i  ft.      1  mile  =  320  rods  =  5280  ft. 
1  meter  =  10  decimeters  =  100  cm.  =  1000  mm. 
1  km.  =  1000  meters. 

Angles:    straight  angle,  right  angle,  acute  angle,  obtuse 
angle,  reflex  angle,  perigon. 
Angle  measure:  360  degrees  make  a  perigon. 

2  TT  radians  make  a  perigon. 

The  letter  ir  denotes  the  number  of  times  the  diameter  of 
a  circle  is  contained  in  the  circumference;  2  tt  is  the  number 
of  times  that  the  radius  is  contained  in  the  circumference. 

c  =  ird  =  2Trr. 

TT  =  3.14159  +  •  •  •  =  W  approximately. 
One  straight  angle  =  180  degrees  =  tt  radians. 

One  right  angle       =    90  degrees  =  ^  radians. 

The  sum  of  the  angles  of  a  triangle  is  two  right  angles. 
The  sum  of  the  angles  of  a  quadrilateral  is  four  right 
angles. 

Areas 


Rectangle  or  Parallelogram 

Triangle 

Circle 

a  =  hh. 

a  =  ihh. 

a  =  wr^. 

'  =  r 

'-¥■ 

a  =  icr. 

"'i- 

-¥• 

II.  §17]  MEASUREMENT  OF  VOLUME  27 

Volumes 

Rectangiilar  Parallelopiped  Cylinder 

Volume  =  abc.  Volume  =  xr^/i. 

Cone  Sphere 

Volume  =  J  Tr%,  Volume  =  |  tt^. 


CHAPTER  III 


100 


212 


MEASUREMENTS  CONTINUED  —  TEMPERATURE, 
WEIGHT   AND   DENSITY,   FORCE 

18.  Measurement  of  Temperature.  A  thermometer  is  an 
instrument  for  measuring  temperature.  It  consists  of  a 
small  bulb  or  reservoir  filled  with  mercury  and 
connecting  with  a  very  narrow  vertical  tube. 
When  the  mercury  in  the  bulb  is  heated  it 
expands  and  rises  into  the  tube.  So  the  height 
of  the  mercury  in  the  tube  indicates  the  degree 
of  heat  to  which  the  bulb  is  exposed. 

The  Centigrade  Scale  of  Temperature.  Im- 
merse a  thermometer  in  melting  ice  and  mark 
the  point  on  the  tube  where  the  mercury  stands 
zero.  This  is  called  the  freezing  point.  Next 
immerse  in  boiling  water  and  mark  the  new 
point  100.  This  is  called  the  boiling  point. 
Divide  the  space  between  zero  and  100  into  100 
equal  parts.  Each  such  part  is  called  one  de- 
gree Centigrade.  The  graduations  are  extended 
below  the  zero  point  to  indicate  temperatures 
below  zero. 

The  Fahrenheit  Scale  of  Temperature.     Pro- 
ceed as  above,  except  that  the  freezing  point 
is  marked  32  and  the  boiling  point  212.    Divide 
the  space  between  these  into  180  equal  parts. 
Each  such  part  is  called  one  degree  Fahrenheit. 

The  adjacent  figure  shows  two  thermometers,  one  gradu- 
ated Centigrade,  the  other  Fahrenheit.  Both  scales  may 
be  placed  on  the  same  thermometer. 

28 


so 


III.  5 19]  MEASUREMENT  OF  WEIGHT  29 

Exercises. 

1.  Immerse  two  thermometers,  one  graduated  C.  and  the 
other  F.,  or  one  thermometer  with  both  scales  on  it,  in  a  cup 
of  cold  water  and  record  the  reading  on  each  scale.  Next 
immerse  in  a  cup  of  hot  water  and  record  the  readings  on 
each.  From  these  readings  find  how  many  degrees  Fah- 
renheit equal  one  degree  Centigrade.  Write  out  each  step 
in  full  as  in  the  example  worked  out  on  p.  16. 

2.  How  many  degrees  Fahrenheit  are  equal  to  100  de- 
grees Centigrade?  From  this  find  how  many  degrees  Fahren- 
heit equal  one  degree  Centigrade. 

3.  When  the  Fahrenheit  scale  reads  70°,  what  is  the 
Centigrade  reading  ? 

4.  When  the  Centigrade  scale  reads  35°,  what  is  the 
Fahrenheit  reading? 

6.  Let  F  denote  the  Fahrenheit  reading,  and  C  denote 
the  corresponding  Centigrade  reading.  Find  C  when  F  = 
86°;  68°;  50°;  41°;  14°.  Fmd  F  when  C  =  10°;  20°;  30°; 
70°;  10°  below  zero. 

6.  Show  that  for  all  temperatures  above  freezing 

JP'  =  32  +  I  C; 
and  that 

C  =  i  (jP  -  32). 

7.  At  what  temperature  Centigrade  is  the  Fahrenheit 
reading  equal  to  three  times  the  Centigrade  reading? 

19.  English  Units  of  Weight.  The  weight  of  a  body  is 
measured  by  the  pull  of  the  earth  upon  that  body.  When 
we  say  that  a  body  weighs  five  pounds,  we  mean  that  the 
earth  pulls  on  it  with  five  times  the  pull  on  a  one-pound 
weight. 

Now  what  is  a  one-poimd  weight?  It  is  the  weight  of  a 
certain  piece  of  platinum  which  is  carefully  preserved  by 
the  British  government,  and  used  to  test  other  pound 
weights. 


30  ALGEBRA  — FIRST  COURSE  [111,5  20 

When  this  weight  is  hung  on  a  spring  balance,  the  place 
to  which  the  pointer  moves  is  marked  1.  Any  other  weight 
which  pulls  the  pointer  to  the  same  place  is  then  also  a 
pound.  By  hanging  on  two  such  weights,  then 
three,  and  so  on,  each  time  marking  the  place  where 
the  pointer  stops,  we  get  a  spring  balance  gradu- 
ated in  pounds.  To  weigh  a  body  we  need  only 
hang  it  on  the  balance  and  notice  where  the  pointer 
stops. 

Based  on  the  pound  we  have  the  English  units  of 
weight  as  follows: 

1  pound  (Avoirdupois)  =16  ounces  =  7000  grains. 
100  pounds  =  1  hundredweight. 
2000  pounds  =  1  ton. 
2240  pounds  =  1  long  ton. 

There  is  another  pound,  called  the  Troy  pound,  in  less 
common  use.     It  is  divided  into  12  ounces  and  5760  grains. 

We  shall  deal  only  with  the  pound  avoirdupois. 
Exercises. 

1.  How  many  ounces  in  2  lb.?  In  3|  lb.?  In  4J  lb.? 
In  n  lb.?  In  (a  +  h)  lb.?  How  many  grains  in  each  of 
these? 

2.  How  many  pounds  in  40  oz.?  In  75  oz.?  In  m  oz.? 
In  Qi  +  h)  oz.?     How  many  grains  in  each  of  these? 

3.  How  many  pounds  in  10,000  gr.?  In  1400  gr.?  In 
r  gr.?     In  (c  —  d)  gr.? 

4.  How  many  grains  in  a  lb.  +  6  oz.?  How  many  ounces 
in  r  lb.  +  ^  gr.? 

20.  Metric  (or  French)  Units  of  Weight.  In  France  the 
unit  of  weight  is  the  weight  of  a  cubic  centimeter  of  water 
at  a  temperature  of  4  degrees  Centigrade.  It  is  called  a 
gram.  This  is  a  rather  small  unit,  for  it  takes  nearly  500 
grams  to  make  a  pound;  therefore  a  larger  unit,  namely 
the  kilogram  or  1000  grams,  is  more  commonly  used.    A 


Ill,  §  21] 


MEASUREMENT  OF  WEIGHT 


31 


kilogram  equals  about  2.2046  pounds.  A  half  kilo  would 
then  be  a  little  more  than  one  pound,  and'*this  is  the  official 
unit  which  takes  the  place  of  the  English  pound. 

For  deUcate  weighings,   such  as  are  needed  in  physics 
and  chemistry,  the  gram  is  divided  into  smaller  units  thus: 
one  tenth  of  a  gram  =  a  decigram; 
one  hundredth  of  a  gram  =  a  centigram; 
one  thousandth  of  a  gram  =  a  milligram. 

Exercises. 

1.  How  many  decigrams  in  10  grams?  In  1.5  grams? 
In  g  grams?     In  {m  +  n)  grams?    How  many  centigrams? 

2.  How  many  milligrams  in  5  decigrams?  In  3  centi- 
grams? In  k  decigrams?  In  k  decigrams  +  I  milligrams? 
How  many  grams  in  each  of  these? 

21.  The  Beam  Balance.  For  accurate  weighing  the 
spring  balance  is  replaced  by  the  beam  balance.  This  is 
constructed  on  the  following  principle. 


n 


n 


When  a  beam  is  supported  at  its  middle  on  a  knife  edge 
and  equal  weights  are  placed  at  equal  distances  from  the 
knife  edge,  the  beam  will  balance. 


For   convenience   a  scalepan  is  usually  fastened  to  the 
beam  at  each  end;   in  the  druggists'  scales  the  pans  rest  on 


32  ALGEBRA  —  FIRST  COURSE  tiii.§22 

top  of  the  beam;    in  the  chemical  balance  the  pans  are 
hung  from  the  beam  by  wires. 

Exercises. 

1.  In  one  pan  of  a  beam  balance  place  a  weight  of  one 
pomid;  how  many  grams  must  be  placed  in  the  other  pan 
to  balance?    How  many  grams  in  a  pound? 

2.  Place  a  piece  of  iron  or  other  substance  in  one  scale- 
pan;  balance  it  first  by  English  units,  then  by  metric  units, 
noting  the  amount  of  each.  From  your  notes  calculate  the 
number  of  grams  in  an  ounce.  Write  out  the  solution  in 
full  as  in  the  example  worked  out  on  p.  16. 

Repeat  this  experiment  several  times  with  different 
amounts  of  material. 

3.  A  certain  weight  is  balanced  by  q  lb.  and  r  oz.  How 
many  grams  does  it  weigh? 

22.  The  Lever.  A  beam  placed  on  a  knife  edge  is  usu- 
ally called  a  lever.  The  ordinary  ''see-saw"  is  a  rough  sort 
of  lever.  The  knife  edge  is  called  the  fulcrum;  the  parts  of 
the  beam  on  each  side  of  the  fulcrum  are  called  the  arms  of 
the  lever. 

If  two  boys,  one  considerably  heavier  than  the  other, 
were  to  play  see-saw,  would  they  sit  at  equal  distances  from 
the  fulcrum?    Why  not?    Which  one  would  sit  farther  out? 

This  illustrates  the  following  rule,  called  the  'principle  of 
the  lever: 


n 

m' 

r 

"^         d, 

d2 

*1 

Rule.  In  order  that  a  weight  Wi,  placed  at  a  distance  di 
from  the  fulcrum,  shall  balance  a  weight  W2  placed  at  a  dis- 
tance dk  from  the  fulcrum,  we  must  have 

Fi  X  c^i  =  TF2  X  d^. 


III.  §  221  MEASUREMENT  OF  WEIGHT  33 

Exercises. 

1.  State  the  last  equation  in  words. 

2.  If  a  boy  weighing  75  lb.  sits  6  feet  from  the  fulcrum, 
where  should  a  boy  weighing  100  lb.  sit  to  balance  the 
beam?  Give  neat  solution,  letting  d  stand  for  the  number 
of  feet  in  the  required  distance. 

3.  How  far  from  the  fulcrum  must  a  12-lb.  weight  be  placed 
to  balance  a  30-lb.  weight  placed  5  feet  from  the  fulcrum? 

4.  A  weight  of  200  grams  is  placed  25  centimeters  from 
the  fulcrum.  How  far  from  the  fulcrum  must  a  weight  of 
half  a  kilogram  be  placed  to  balance? 

When  several  weights  are  placed  at  various  distances  to 
one  side  of  the  fulcrum,  and  other  weights  at  various  dis- 
tances to  the  other  side  of  the  fulcrum,  the  beam  will  balance 
when  the  sum  of  the  products  formed  by  multiplying  each 
weight  on  one  side  by  its  distance  from  the  fulcrum  is  equal  to 
the  sum  of  such  products  from  the  other  side. 

Example.  Weights  of  5,  10  and  3  lb.  respectively  are  placed  at  the 
distances  shown  in  the  figure.  What  weight  W  placed  8  feet  from  the 
fulcrum  will  balance  the  beam? 


e,,  10  lb.  Wlb. 


B. 


m. 


L « ^ 


-io >|4-— -« J 

Solution:  According  to  the  last  rule  we  must  have 

5X10  +  10X6  =  3X2  +  TFX8. 
That  is,  110  =  6  +  8  W, 

or  104  =  SW. 

Therefore,  Tf  =  13  lbs. 

Check  this  answer. 

Exercises.     Find  what  is  necessary  for  balance  in  each  of 
the  following  cases. 


34  ALGEBRA  — FIRST  COURSE  [ill.  §23 

To  left  of  fulcrum  To  right  of  fulcrum 

1.  TF  =  3  lb.    7  lb.;  5  lb.    ?  lb. 

d=      4  ft.     6  ft.;  8  ft.    2  ft. 

2.  W  =      8oz.    5oz.;  .      ?  oz.    10  oz. 

d  =      7  in.           4  in. ;  13  in:.     5  in. 

3.  IT  =  100  gm.   75  gm.;  60  gm.   20  gm. 

d  =      5  cm.    8  cm.;  ?  cm.   10  cm. 

4.  Tf  =   2  kgm.   5  kgm.  4  kgm. 


d  =    25  cm.   15  cm.    ?  cm.;       30  cm. 
b.  W  =      2ilb.;  3Hb.   1.4  lb.   21b. 

d=    25  in.;  6  in.     ?  in.   9|  in. 

23.  Density  —  Specific  Gravity. 

Exercise  1.  Determine  the  weight  of  a  solid  rubber  ball 
in  grams.  Determine  the  volume  of  the  ball  by  dropping 
it  into  water  in  a  graduated  beaker  and  noticing  the  amount 
the  water  rises.  Express  this  volume  in  cubic  centimeters. 
How  many  grams  does  this  amount  of  water  weigh?  Divide 
the  weight  of  the  ball  by  the  weight  of  the  same  volume  of 
water.     The  quotient  is  called  the  density  of  the  ball. 

Definition.  The  density  of  any  substance,  when  water  is 
taken  as  the  standard,  is  the  quotient  obtained  by  dividing 
the  weight  of  a  given  volume  of  that  substance  by  the 
weight  of  an  equal  volume  of  water.  The  density  of  a 
liquid  is  also  often  called  "specific  gravity.'' 

Exercise  2.     Determine  the  density  of  iron,  wood,  stone, 
and  glass  by  repeating  Exercise  1  with  pieces  of  each  of 
these  substances.     Make  a  table  of  your  results. 
Other  Exercises. 

3.  Find  the  density  of  a  solution  made  by  dissolving  10 
gm.  common  salt  in  20  gm.  water.  (Divide  the  weight  of 
the  solution  by  the  weight  of  an  equal  volume  of  water.) 

4.  Find  the  density  of  a  solution  made  by  dissolving 
w  grams  of  salt  in  n  grams  of  water. 

6.  How  many  grams  of  salt  should  be  dissolved  in  25  cc. 
water  to  make  a  solution  whose  density  is  1.4? 

6.  To  10  cc.  of  a  liquid  of  sp.  gr.  1.2  are  added  20  cc.  of 
a  liquid  of  sp.  gr.  0.8.     What  is  the  sp.  gr.  of  the  mixture? 


in.  §  2i] 


SUMMARY 


35 


Table  of  Densities 


Aluminum 2.6 

Brass 8.4 

Charcoal 1.6 

Copper 8.8 

Cork 14r-.24 


Diamond. . . . 

Glass 

Gold 

Gutta  Percha , 

Iron 

Ivory 


Alcohol . 
Benzine 
Ether.  . 


3.53 
4r-4.5 
19.3 

0.97 

7.8 
1.8 


Lead 11.4 

Marble 2.6 

Mercury 13 . 6 

Nickel 8.7 

Paraffin 0.89 

Platinum 21.5 

Silver 10.5 

Tin 7.3 

Zinc 7.2 

Water 1.0 


Liquids 

0.81        Glycerine 1.27 

0.90        Turpentine 0.88 

0.73        Water 1.00 


24.   Summary. 


Temperature 


Centigrade  scale:  freezing  =    0  degrees; 

boiling  =  100  degrees. 
Fahrenheit  scale:  freezing  =    32  degrees; 

boiling  =  212  degrees. 


English 


Units  of  Weight 


Metrics 


Grain                   (gr.) 

Milligram  (mgm.) 

Ounce                  (oz.) 

Centigram  (cgm.) 

Pound                  Gb.) 

Decigram    (dgm.) 

Hundredweight  (cwt.) 

Gram           (gm.) 

Ton 

Kilogram     (kgm.) 

The  Lever.     To  balance  a  lever:    Multiply  each  weight 
on  one  side  of  the  fulcrum  by  its  distance  from  the  fulcrum 


36  ALGEBRA  — FIRST  COURSE  [iii.§25 

and  form  the  sum  of  these  products;  do  the  same  with  the 
weights  on  the  other  side  and  form  the  sum  of  these  products; 
the  two  sums  must  be  equal. 

The  density,  or  specific  gravity,  of  a  substance  is  the 
quotient  obtained  by  dividing  the  weight  of  a  given  volume 
of  that  substance  by  the  weight  of  an  equal  volume  of  water. 

25.   Questions  and  Problems  for  Review. 

1.  What  are  the  units  used  to  measure  angles?  Describe 
each. 

2.  What  are  the  common  English  units  used  to  measure 
lengths,  areas,  volumes,  and  weights? 

3.  What  are  the  French  or  metric  units? 

4.  Define  density. 

5.  Define  the  meaning  of  the  words  "lever''  and  "ful- 
crum.'' 

6.  What  is  the  sum  of  the  angles  of  a  triangle  in  degrees? 
In  radians?  What  is  the  sum  of  the  angles  of  a  quadri- 
lateral? 

7.  On  cross-section  paper  draw  a  rectangle  of  any  con- 
venient width  and  height.  Draw  a  triangle  of  the  same 
width  and  height  as  the  rectangle.  ,  By  counting  squares 
verify  the  formula 

Area  of  triangle  =  J  (area  of  rectangle). 

Express  this  formula  in  letters,  letting  t  stand  for  the  area 
of  the  triangle  and  r  for  the  area  of  the  rectangle. 

8.  Draw  a  circle;  also  draw  a  triangle  whose  base  equals 
the  perimeter  of  the  circle  and  whose  height  equals  the  radius 
of  the  circle.  By  counting  squares  compare  the  areas  of  the 
two  figures  as  well  as  possible.  If  a  stands  for  the  number  of 
square  units  in  the  area  of  the  circle,  c  for"  the  number  of 
linear  units  in  the  circumference,  and  r  for  the  number 
of  linear  units  in  the  radius,  do  your  numbers  fit  the  equa- 
tion 


III.  §  25]  PROBLEMS  37 

We  have  also  found  experimentally  that 

c  =  2  Trr. 
Usmg  these  two  equations,  show  that 
a  =  Trr 2     (p.  23). 

9.  If  one  angle  is  2  times  the  size  of  another  angle,  and 
the  sum  of  the  two  angles  is  x  radians,  what  is  the  number 
of  radians  hi  each  angle?  Draw  these  angles,  after  you 
have  found  the  size  of  each. 

Solution: 

Let  a  =  the  number  of  radians  in  the  first  angle. 

Then  2  a  =  the  number  of  radians  in  the  second  angle, 

and  3  o  =  the  number  of  radians  in  the  sum  of  the  two  angles. 

But  X  =  the  number  of  radians  in  both  angles. 

Therefore  3  a  =  tt,  since  3  a  and  x  stand  for  the  same  number  of 
radians. 

Then  a  =  5 »  ^^®  number  of  radians  in  the  first  angle, 

and  2  a  =  -^ ,  the  number  of  radians  in  the  second  angle, 

o 

Therefore  the  first  angle  contains  5  radians,  and  the  second  angle 

o 

contains  -^  radians. 

Check:  The  next  step  is  to  check  the  correctness  of  our  answers. 
To  do  this  we  go  back  to  the  original  statement  of  the  problem  and  see 
whether  every  statement  is  fulfilled.  To  begin  with,  the  one  angle 
must  be  twice  as  large  as  the  other. 

-K-  radians  =  2X5  radians. 

Also,  the  sum  of  the  two  angles  must  be  ir  radians. 

5  radians  +  -^  radians  =  ir  radians. 

6  6 

Therefore  our  answers  are  correct. 

10.  If  one  angle  is  2  times  the  size  of  another  angle,  and 
the  sum  of  the  two  angles  is  180  degrees,  what  is  the  number 
of  degrees  in  each  angle?    Draw. 


38  ALGEBRA  —  FIRST  COURSE  lin.§25 

11.  Take  your  answers  to  exercise  9  and  solve  to  find  the 
number  of  degrees  in  each  angle.  Make  use  of  letters  in 
your  solution.  How  do  the  answers  of  exercises  10  and  11 
compare? 

12.  The  sum  of  two  angles  is  f  tt  radians,  and  one  of  them 
is  3  times  the  size  of  the  other.  What  is  the  number  of 
radians  in  each?     Draw. 

13.  State  exercise  12  expressing  the  sum  in  degrees.  Solve 
and  show  agreement  of  answers. 

14.  The  sum  of  three  angles  is  |  radians.  The  first  is  J 
of  the  size  of  the  second,  and  the  third  is  J  of  the  size  of  the 
first.  What  is  the  number  of  radians  in  each  angle?  Draw 
these  angles. 

16.  State  exercise  14,  expressing  the  sum  in  degrees  and 
compare,  as  in  exercise  13. 

16.  The  length  of  a  rectangle  is  3  times  its  width.  The 
length  of  the  line  bounding  it  (called  the  perimeter)  is  372  mm. 
What  is  the  length  of  the  rectangle?  The  width?  Draw  the 
rectangle. 

17.  State  exercise  16,  using  the  inch  unit  to  express  the 
combined  length  of  the  three  lines.  Solve,  draw,  and  com- 
pare. 

18.  Of  five  lines  the  first  is  2J  times  the  second,  the  third 
is  I  as  long  as  the  first,  the  fourth  is  as  long  as  the  sum  of 
the  first  and  third,  and  the  fifth  is  |  as  long  as  the  sum  of 
the  first  and  fourth.  What  is  the  length  of  each  line  if  the 
combined  length  is  27  millimeters? 

19.  If  one  of  the  angles  of  a  triangle  is  J  as  large  as  an- 
other, and  the  third  is  j  of  the  sum  of  the  first  and  second, 
what  is  the  number  of  radians  in  each  angle?  Draw  such  a 
triangle. 

20.  If  the  first  of  three  angles  of  a  triangle  is  ?  as  large  as 
the  second  and  the  third  is  f  as  large  as  the  first,  what  is 
the  number  of  radians  in  each  angle?    Draw  such  a  triangle. 

21.  The  first  of  the  three  angles  of  a  triangle  is  half  as  large 


III.  1 25]  PROBLEMS  39 

as  the  second,  and  the  third  is  equal  to  the  sum  of  the  first 
and  second.  What  is  the  number  of  degrees  in  each  angle? 
Draw. 

22.  The  first  angle  of  a  quadrilateral  is  f  as  large  as  the 
second,  the  third  is  equal  to  the  first,  and  the  fourth  is  equal 
to  the  second.  What  is  the  number  of  radians  in  each  angle 
of  the  quadrilateral?     Draw  such  a  quadrilateral. 

23.  One  of  the  four  angles  of  a  quadrilateral  is  ^  as  large 
as  another,  a  third  is  equal  to  the  sum  of  the  two,  and  the 
fourth  is  equal  to  the  difference  between  the  two.  What  is 
the  number  of  degrees  in  each  angle?  Draw  such  a  quadri- 
lateral. 

24.  In  the  following  state  the  problem  in  good  English, 
using  the  word  instead  of  the  letter  which  stands  for  that 
word.  Solve  by  substituting  the  number  instead  of  the 
letter  in  the  formula  on  p.  26,  and  answer  the  question 
asked  in  your  problem.     Illustrate  each  problem. 

For  the  rectangle  or  parallelogram: 

(a)  h  =  7,     h  =  2,     a  =?     (See  formulas  on  p.  26.) 

Illustration.  If  a  rug  is  7  feet  long  and  2  feet  wide,  what  is  the  num- 
ber of  square  feet  in  the  rug? 

Solution:  Formula,  a  =  hh. 

Since  6  =  2    and    h  =  7, 

a  =  2  .  7  =  14. 

Therefore  there  are  14  square  feet  in  the  area  of  the  rug. 

(6)  a  =  100,  6=8,  h  =2 

(c)  a  =  57,  /i=3,  b  =? 

id)  a  =  TV,  b=h  h=? 

(e)  6=53f,  A  =  7f,  a  =? 

For  the  triangle: 

(/)     6  =  7.5,  /i=40,  a=? 

(g)    a  =  286,  h  =459,  h  =? 

{h)    a  =  2h  /i  =  lf,  h  =? 


40  ALGEBRA  —  FIRST  COURSE  llll,§25 


)                          ALGEBRA - 

-FIRST  COURSE 

For  the  circle: 

(0     r  =  110, 

c  =?             a=? 

0*)     c  =  3.5, 

r  =?            a=? 

(k)    c?  =  .42, 

c  =?            a=? 

(I)     a  =  49  TT, 

r =?            c  =? 

25.  If  a  certain  sum  of  money  is  put  at  simple  interest  at 
a  certain  rate,  for  a  certain  time,  you  have  learned  from  your 
arithmetic  that 

interest  =  principle  X  rate  X  time. 

Using  the  first  letter  of  each  word  as  we  have  done  in 
previous  examples, 

i  =  p  'T  *t. 

(a)  What  does  p  equal  in  terms  of  i,  r,  Vt  What  does  t 
equal  in  terms  of  p,  i,  r?     What  does  r  equal  in  terms  of  p,  i,  Vt 

(b)  If  a  sum  of  200  dollars  is  placed  at  simple  interest  at 
6%  for  3  years,  what  is  the  interest  due?  Solve  by  using 
formula  and  answering  question  asked. 

(c)  The  interest  on  $1200  placed  at  the  rate  of  5%  was 
$240.     What  was  the  time  of  the  note? 

(d)  At  what  rate  of  interest  must  $1400  be  placed  in 
order  that  it  bring  $294  interest  by  the  end  of  three  years? 

26.  An  ivory  ball  and  an  India-rubber  ball  of  the  same 
size  together  weigh  60  grams.  Ivory  is  twice  as  heavy  as 
India  rubber.     What  is  the  weight  of  each  ball? 

27.  A  solution  of  alum  and  water  weighs  456  grams. 
Alum  weighs  1.7  times  as  much  as  water.  How  many 
grams  of  alum  are  there  in  the  solution? 

28.  Marble  is  1.5  times  as  heavy  as  ivory.  9  marble  balls 
weigh  46.8  grams  more  than  7  ivory  balls.  What  is  the 
weight  of  each  ball? 

29.  A  brass  ornament  plated  with  gold  weighs  50  grams. 
There  are  5  times  as  many  cubic  millimeters  of  brass  as 
there  are  of  gold,  and  one  cubic  millimeter  of  gold  weighs 


Ill,  §25]  PROBLEMS  41 

.0193  grams,  and  one  cubic  centimeter  of  brass  weighs  8.4 
grams.  How  many  cubic  millimeters  of  each  metal  are 
there  in  the  ornament? 

30.  If  we  know  the  velocity  of  a  moving  body,  and  the 
length  of  time  it  moves,  how  may  we  find  the  distance  it 
moves? 

Let  V  stand  for  the  velocity,  t  for  the  time,  and  d  for  the 
distance;  state  the  answer  to  the  last  question  in  letters. 

State  the  formula  for  t  in  terms  of  d  and  v.  State  the  for- 
mula for  V  in  terms  of  d  and  L  Make  problems  for  each  of 
these  cases. 


CHAPTER  IV 
GRAPHIC  REPRESENTATION  OF  QUANTITY 

26.  The  Amount  of  any  Quantity  Represented  by  the 
Length  of  a  Line.  The  result  of  the  measurement  of  any 
quantity  may  be  expressed  by  a  number  which  tells  us  how 
many  units  or  parts  of  units  of  measure  are  contained  in  the 
quantity  measured.  We  may  also  represent  the  amount 
of  our  quantity  by  the  length  of  a  line;  this  is  especially  use- 
ful when  several  quantities  are  to  be  compared,  because  the 
eye  takes  in  at  a  glance  the  various  lengths  of  lines. 

Example  1.  On  your  section  paper  let  one  division  of  the  line  rep- 
resent 1  pound;  then  four  divisions  will  represent  4  pounds;  six  divisions 
will  represent  6  pounds;  and  so  on. 


lib. 


klh. 


6  lb. 


Example  2.     Let  two  divisions  represent  a  distance  of  1  mile.     Then 
4  divisions  will  represent  2  miles;  7  divisions  will  represent  3|  miles. 


1  mi. 


2  mi. 


sVs  mi 


42 


IV.  §26]     GRAPHIC  REPRESENTATION  OF  QUANTITY       43 

On  the  scale  of  ^  inch  to  the  mile,  how  wide  would  a  map  of  the 
United  States  be?     (The  distance  east  and  west  is  about  3000  miles.) 

How  wide  would  a  map  of  your  state  be?     How  high? 

Example  3.  If  a  line  1  inch  long  represents  10,000  people,  then  a 
line  2 1  inches  long  represents  25,000  people,  a  line  |  inch  long  repre- 
sents 2500  people. 


10,000 


^S,000 


2500 


Example  4.  To  compare  unequal  quantities  of  the  same  kind,  it  is 
very  convenient  to  represent  them  by  parallel  hnes  drawn  under  each 
other.     This  is  illustrated  below. 


Lengths  of  rivers. 

Mississippi 4300  miles 

Amazon 3300  miles 

Nile 3400  miles 

Volga 2400  miles 


M 


i^ssi: 


£L 


NMe 


Vdlga 


Example  5.  We  may  also  draw  the  lines  representing  the  values  of 
quantities  perpendicularly.  Thus  in  the  following  diagram  the  heavy 
vertical  lines  represent  the  relative  heights  of  four  mountain  peaks. 


44 


ALGEBRA  —  FIRST  COURSE 


[IV.  §  26 


Heights  of  mountains. 

Pike's  Peak 14,000  feet 

Mt.  Blanc 16,000  feet 

Chimborazo 20,000  feet 

Mt.  Everest 29,000  feet 

Exercises.  In  the  following 
select  an  appropriate  unit  and 
represent  by  drawings. 

1.  Draw  lines  whose  lengths 
shall  represent  3  lbs. ;  2§  lbs. ;  2f 
lbs.;     20  oz.;  50  oz. 

2.  On  the  scale  of  one  unit  of 
division  on  your  section  paper  to 
the  foot,  what  lengths  are  repre- 
sented by  the  following:  3  divi- 
sions; 6  divisions;  12  divisions; 
36  divisions;  90  divisions.  Draw 
lines  representing  5  feet;  3|  feet; 
1§  yards;  J  rod. 

Make  a  diagram  of  the  floor 
of  a  room  whose  dimensions  are 
10  feet  by  15  feet,  on  a  scale  of 
one  division  to  five  feet. 

3.  Selecting  an  appropriate  length  to  represent  one  day, 
tell  what  durations  of  time  the  following  lengths  would 
represent?     7;  20;  jV;  ^is- 

How  long  a  line  would  be  required  to  represent  10  days? 
one  year?  Draw  lines  to  represent  4  hours;  75  minutes; 
1  hour  and  40  minutes. 

4.  Express  by  drawings  the  yearly  interest  on  $100  at 
1%;  at  2%;  at  4%;  at  7%;  at  10%. 

On  the  same  scale  represent  the  yearly  interest  on  $200 
at  1%;  on  $400  at  1%;  on  $300  at  2%. 

5.  Selecting  a  length  to  represent  100,000  people,  represent 
by  lengths  of  lines  the  population  of  the  following  cities: 


so  000 

ft. 

25  000 

20  000 

15  000 

<w 

10  000 

o 

^ 

•n 

g 

S 
s 

^ 

5000 

? 

D 

g 

0 

IV.  §26]     GRAPHIC  REPRESENTATION  OF  QUANTITY        45 


Pittsburg 
375,000 


Cleveland 
480.000 


City:  Milwaukee    Buffalo    Baltimore 

Population:  320,000      400,000      550,000 

6.  By  measurement  on  a  map,  using  the  scale  given  on 
the  map,  find  the  distances  between  various  cities.  Make  a 
table  of  the  results  and  then  show  the  distances  by  lines. 

7.  From  the  following  diagram  read  off  the  population 
of  the  United  States  for  the  various  dates. 


■-BvO 

mo- 


1890- 


•  -mo- 


50 
PopulUtion  in  million 


100 


8.   From  the  following  diagram  read  off  the  amount  of 
national  debt  per  capita  in  different  years. 


ion  ft 

0 

6 

)  dolU 

irs 

L 

National  debt  per  Capita  in  CT.S. 


46 


ALGEBRA  —  FIRST   COURSE 


IIV.  §  26 


^10. 


r~i 


I      I       5 


-f — -f 1^ 1- 

World's  Production  of  Gold 


9.  From  the  preceding  diagram  read  off  the  world's  pro- 
duction of  gold  in  the  various  years. 

Make  diagrams  to  show  the  data  given  in  the  tables  below. 

10.  Distance  from  New  York  to  Chicago,  910  miles, 

to  Boston,  160  miles, 

to  Cleveland,       580  miles, 
to  Washington,  230  miles. 

11.  Population  of  countries,  1910: 

United  States,     93  milUon  people, 
England,  45  million  people, 

France,  39  million  people, 


IV.  §  26]    GRAPHIC  REPRESENTATION  OF  QUANTITY        47 

Germany,  63  million  people, 

Italy,  34  million  people, 

Japan,  51  million  people. 

12.  Number  of  people  to  the  square  mile,  1910: 

United  States,  26  persons, 

England,  374  persons, 

France,  191  persons, 

Germany,  311  persons, 

Italy,  313  persons, 

Japan,  344  persons. 

13.  Distances  from  the  sun  to  the  four  inner  planets: 

Sun  to  Mercury,  36  million  miles, 

Sun  to  Venus,  67  million  miles, 

Sun  to  Earth,  93  million  miles. 

Sun  to  Mars,  141  million  miles. 

14.  Make  diagrams  showing  various  data.  The  following 
are  suggested. 

Population  of  your  state  for  every  ten  years. 

Amount  of  public  debt  of  your  state  for  several  years. 

Amount  of  crops  —  com,  wheat,  etc. 

Value  of  crops. 

Amount  of  various  manufactures. 

Population  of  leading  cities. 

The  comparison  of  quantity  is  also  expressed  by  the  com- 
parative sizes  of  angles.  The  amount  of  turning  or  circular 
motion  is  quite  as  effective  in  its  power  to  convey  compari- 
sons as  is  the  straight  line.  It  is  not  used  so  extensively, 
because  a  straight  line  is  easier  to  make.  The  following 
pictures  illustrate  its  use. 


48 


ALGEBRA  —  FIRST  COURSE 


[IV,  §  26 


Population  of  Cities 


Speed  Records 


London,  7 . 2  million, 
New  York,  5 . 2  million, 
Paris,  2.8  million, 

Chicago,  2.2  million, 
Tokio,  2.2  million. 


Berlin, 


2.0  million. 


Automobile,  142  miles  per  hour, 
Locomotive,  120  miles  per  hour, 
Aeroplane,  106  miles  per  hour, 
Pidgeon,  86  miles  per  hour, 

Horse,  43  miles  per  hour, 

100  yd.  dash,  21  miles  per  hour. 


CHAPTER  V 
POSITIVE   AND    NEGATIVE   NUMBERS 

27.  Exercises. 

1.  A  boy  strikes  a  ball  at  exactly  opposite  points  with  two 
mallets  at  the  same  time  and  with  the  same  force.  What 
is  the  result  as  to  the  movement  of  the  ball? 

2.  If  he  strikes  first  with  the  right-hand  mallet,  and  then, 
after  the  ball  has  come  to  rest,  with  the  left=hand  mallet 
with  an  equal  force,  what  is  the  result? 

3.  If  he  strikes  with  both  mallets  at  the  same  time,  strik- 
ing with  one  mallet  with  a  force  which  would  send  the  ball 
5  feet  and  with  the  other  mallet  with  a  force  which  would 
send  the  ball  20  feet,  what  is  the  result? 

4.  If  in  Exercise  3  the  boy  strikes  first  with  one  mallet, 
then,  after  the  ball  has  come  to  rest,  with  the  other  mallet, 
what  is  the  result? 

5.  Does  it  make  any  difference  whether  two  forces  act 
at  the  same  time  or  one  after  the  other? 

6.  Two  men  are  driving  a  stake;  one  strikes  with  a  force 
that  sends  the  stake  down  two  inches,  the  other  with  a 
force  that  sends  the  stake  down  three  inches.  What  is  the 
combined  effect? 

7.  Three  boys  are  pulling  a  load  on  a  sled,  one  with  a  force 
of  25  pounds,  another  with  a  force  of  58  pounds,  and  the 
other  with  a  force  of  97  pounds.  With  what  force  is  the 
load  being  pulled? 

8.  Two  small  boys  are  pulling  a  small  wagon  along;  one 
pulls  with  a  force  of  25  pounds,  and  the  other  pulls  with  a 
force  of  34  pounds.     A  boy  comes  up  behind  and  pulls 

49 


50  ALGEBRA  — FIRST  COURSE  [V.  §  28 

with  a  force  of  57  pounds  in  the  opposite  direction.  What 
is  the  result? 

From  the  exercises  given,  and  with  a  httle  further  experi- 
ment and  thought  of  your  own,  you  will  be  thoroughly  con- 
vinced that  forces  are  constantly  acting  against  each  other. 
An  object  cannot  stop  itself ^when  it  is  once  in  motion,  neither 
can  it  change  the  rate  at  which  it  is  moving,  nor  the  direc- 
tion in  which  it  is  moving,  any  more  than  it  can  start  itself 
to  moving.  This  truth  is  known  as  Newton's  first  law  of 
motion. 

28.  Opposite  Qualities.  The  exercises  given  have  dealt 
with  forces.  Almost  all  quantities  with  which  we  deal  may 
bethought  of,  and  commonly  are  thought  of,  as  having  two 
opposite  qualities  which  tend  to  counteract  or  neutralize 
each  other. 

East  is  opposite  to  west;  if  we  go  east  any  distance,  then 
west  the  same  distance,  we  come  back  to  the  starting  point. 

In  the  same  way  we  have : 

North  opposite  to  south; 
Up  opposite  to  down; 
Right  opposite  to  left; 
Forward  opposite  to  backward; 
Profit  opposite  to  loss; 
Assets  opposite  to  liabilities; 
Future  time  opposite  to  past  time; 
Temperature  above  zero  opposite  to  temperature 
below  zero. 

Example  1.  A  man  starts  from  a  certain  town  A  and  goes  10  miles 
due  east;  from  the  point  where  he  now  is  he  goes  15  miles  due  west. 
Where  is  he  then  with  respect  to  the  original  starting  point  A? 

The  answer  is:  5  miles  west  from  A.  We  must  state  not  only  the 
distance  from  A,  but  also  the  direction. 

Example  2.  A  man  has  $10,000  assets  and  $15,000  liabilities.  What 
is  his  financial  status? 

The  answer  is:  He  owes  $5000.  It  is  not  enough  to  say  merely 
$6000.     We  must  also  say  whether  it  is  an  asset  or  a  liability. 


V.§29]  POSITIVE  AND  NEGATIVE  NUMBERS  51 

Example  3.  If  at  noon  the  temperature  is  20  degrees  above  zero,  and 
if  the  temperature  rises  10  degrees  during  the  afternoon,  then  falls  40 
degrees  during  the  night,  what  is  the  temperature  the  next  morning? 

Answer:  10  degrees  below  zero.  It  is  not  enough  to  say  10  degrees. 
We  must  also  say  whether  it  is  above  or  below  zero. 

As  we  have  seen,  forces,  whether  acting  at  the  same  time 
or  at  different  times,  may  act  in  opposite  directions  to  one 
another.  In  mathematics  we  express  the  idea  that  one 
force  acts  in  the  opposite  direction  to  another  force  by  saying 
that  one  is  the  negative  of  the  other,  negative  meaning 
opposite. 

Definition.  Whenever  a  quantity  has  two  opposite  quali- 
ties, we  call  one  of  them  positive  and  the  other  negative. 

Notation.  The  symbol  for  positive  is  +,  and  the  symbol 
for  negative  is  — .  When  no  sign  is  written,  the  +  sign  is 
understood. 

Illustrations. 

If  +  100  dollars  means  gain,  then  —  100  dollars  means  loss. 

If  +  10  feet  means  10  feet  up,  then  —  10  feet  means  10  feet  down. 

If  +  10  days  means  10  days  later,  then  —  10  days  means  10  days 
earlier. 

If  +  m  degrees  means  an  angle  measured  counterclockwise,  then 
—  m  degrees  means  an  angle  measured  clockwise. 

If  +  n  means  count  n  units  forward,  then  —  n  means  count  n  units 
backward. 

Exercise. 

(a)  If  earning  money  is  positive,  what  is  negative? 

(6)  If  going  west  is  positive,  what  is  negative? 

(c)  If  upstream  is  positive,  what  is  negative? 

{d)  If  the  pull  of  gravity  is  positive,  what  is  the  pull  of  a 
balloon? 

29.  Positive  and  Negative  Numbers. 

Definition.  Numbers  which  express  the  measurement  of 
positive  quantity  are  called  positive  numbers.  Numbers 
which  express  the  measurement  of  negative  quantity  are 
called  negative  numbers. 


52  ALGEBRA  —  FIRST  COURSE  [v.  §  30 

Notation.  A  positive  number  is  indicated  by  the  sign  + 
written  before  the  number;  a  negative  number  is  indicated 
by  the  sign  —  written  before  it.  When  no  sign  is  written, 
the  positive  sign  is  understood. 

Thus  if  +  10  means  10  units  of  quantity  of  a  certain 
kind,  then  —  10  means  10  units  of  quantity  of  the  opposite 
kind. 

Let  the  student  give  a  number  of  illustrations. 

Absolute  Value  of  a  Number.  When  we  wish  to  indicate 
merely  the  value  of  a  number,  without  regard  to  its  sign, 
we  use  the  symbol  |  |. 

Thus  I  5  I  =+5,  and  also  |  -  5  |  =+  5. 

30.  Graphic  Representation  of  Positive  and  Negative 
Numbers.  We  have  already  shown  how  we  use  lengths  of 
lines  to  represent  numbers,  and  how  the  comparative  lengths 
of  lines  give  a  clear  illustration  of  comparative  values  of 
numbers.  We  have  dealt,  however,  with  numbers  of  the 
same  quality  or  sign. 

We  now  show  how  to  represent  numbers  of  opposite 
quality,  so  that  our  diagram  will  show  both  value  of  the 
number  and  its  quality. 

Let  us  mark  off  a  series  of  equal  distances  on  an  indefinite 
straight  line.  Mark  one  of  the  points  of  division  zero. 
This  is  the  point  from  which  we  begin  to  count.  Mark  the 
points  to  the  right  of  this  1,  2,  3  and  so  on.  Mark  the  points 
to  the  left,  —  1,  —  2,  —  3,  and  so  on.     Thus 

-6-5-/^-5-2-1         0          12          S         k         5         6 
^-1 I I I I 1 1 1 J 1 I I 1_ 

On  this  scale  we  can  show  numbers  of  opposite  quality. 
Example  1.     Represent  $3500  profit  and  $2500  loss. 

>         I     ^  T        7        ,        !        T        7  . u 

Loss  Profit 

Scale,  1  division  to  1000  dollars. 


V.§3i]  POSITIVE  AND  NEGATIVE  NUMBERS  53 

Example  2.    Represent  4.5  miles  east  and  3  miles  west. 

-U  -3  -2  -]  0  1  2  3  I,  5 

-^ '.      '        '        I        '        ■        '        '  ,     ' 

West  East 

Scale,  1  division  to  the  mile. 

Example  3.    Represent  22  pounds  pull  and  45  pounds  push. 

^30  -^0  -10  0  10  2U  '30  kO  50 

— 1 1 1 1 -t 1 1 \ 1- 

Pull  '  Push 

Scale,  1  division  to  10  pounds. 

Selecting  a  suitable  unit,  make  diagrams  of  the  following: 

4200  births,  2500  deaths. 

15  degrees  rise,  35  degrees  fall. 

In  all  of  these  diagrams  the  amount  of  the  quantity  is 
expressed  by  the  number  and  is  represented  by  a  line  of 
the  proper  length;  opposite  qualities  are  expressed  by  lines 
dra^vn  in  opposite  directions.  In  each  case  one  of  the  lines 
represents  a  positive  number  and  the  other  line  a  negative 
number.  That  is,  positive  and  negative  numbers  are  repre- 
sented by  oppositely  directed  lines. 

31.   Summary. 

Many  quantities,  such  as  force,  distance,  time,  and  tem- 
perature, admit  of  the  notion  of  opposite  qualities. 

In  mathematics  opposite  qualities  are  usually  distinguished 
by  the  words  positive  and  negative. 

Measurements  of  quantities  are  expressed  by  numbers; 
measurement  of  positive  quantities  by  positive  numbers; 
measurement  of  negative  quantities  by  negative  numbers. 

The  symbol  +  indicates  positive. 

The  symbol  —  indicates  negative. 

The  absolute  value  of  a  number  is  expressed  by  |     |. 

Geometrically,  the  measurements  of  quantities  of  opposite 
qualities  are  expressed  by  oppositely  directed  Hnes. 


54  ALGEBRA  —  FIRST  COURSE  [V,§31 

Exercises.     Represent  by  diagrams  each  of  the  following  pairs 
of  quantities: 

1.  10  miles  north;  15  miles  south. 

2.  15  pounds  push;  25  pounds  pull. 

3.  60  degrees  above  zero;  20  degrees  below  zero. 

4.  $2000  income;  $1500  expenditure.    • 

5.  If  a  man  earns  $150  a  month  and  spends  $100  a  month, 
show  his  total  earnings  and  expenditures  in  one  year. 

6.  If  a  tank  is  filled  by  a  pipe  which  flows  100  gallons  a 
minute  and  emptied  by  a  pipe  which  flows  60  gallons  a  minute, 
represent  the  total  inflow  and  outflow  in  5  minutes  when  both 
pipes  are  open. 


CHAPTER  VI 
ADDITION   AND    SUBTRACTION 

32.  Notation.  The  signs  of  addition,  subtraction,  multi- 
plication and  division  are  the  same  as  in  arithmetic. 

To  indicate  that  two  numbers  a  and  b  are  to  be  added  we 
write  a  +  6;  this  means  that  h  is  to  be  added  to  a.  Now 
the  numbers  a  and  h  may  be  both  positive;  one  positive,  the 
other  negative;  or  both  negative.     Thus: 

(+  3)  +  (+  4)  means  add  positive  4  to  positive  3; 
(+  3)  +  (—  4)  means  add  negative  4  to  positive  3; 
(—  3)  +  (+  4)  means  add  positive  4  to  negative  3; 
(—  3)  +  (—  4)  means  add  negative  4  to  negative  3. 

When  a  number  is  positive  we  usually  do  not  write  its 
sign;  thus  3  means  +  3,  and  3  +  4  means  (+  3)  +  (+  4). 

We  shall  now  illustrate  geometrically  what  is  meant  by 
adding  two  numbers  together,  whatever  their  signs  may  be. 
We  shall  represent  each  number  by  a  directed  line;  we  then 
add  these  directed  line  segments. 

33.  Addition  of  Directed  Line  Segments.  Let  us  call 
the  point  from  which  a  line  segment  starts  the  initial  point 
of  the  segment,  -and  the  point  where  it  ends  the  final  point 
of  the  segment;  the  final  point  will  be  marked  by  an  arrow- 
head to  show  the  direction  of  the  line. 

Two  directed  line  segments  are  added  by  placing  the  ini- 
tial point  of  the  line  we  are  adding  on  at  the  final  point  of 
the  line  we  are  adding  to,  keeping  each  line  in  its  original 
direction.  The  line  drawn  from  the  initial  point  of  the  line 
we  are  adding  to,  to  the  final  point  of  the  line  we  are  adding 
on,  is  the  sum  of  the  two  lines.     The  direction  of  the  sum 

55 


56  ALGEBRA  — FIRST  COURSE  IVI.  §  33 

is  always  from  the  initial  point  of  the  hne  added  to,  to  the 
final  point  of  the  line  added  on. 

This  is  illustrated  in  the  following  figures;  here  lines  ex- 
tending toward  the  right  are  called  positive,  toward  the 
left  negative. 

Illustration  I. 

To  add  the  line  RS  to  the  line  MN. 

J2  +  51  K 4^ ^ 


^ 


Starting  at  the  initial  point,  draw  the  line  MN.  Now  add  the  line 
RS  by  placing  the  point  R  on  the  point  N,  taking  care  that  each  line  is 
kept  in  its  original  direction. 


E ± iL 


The  distance  MS  is  the  sum.    It  is  a  positive  line. 
Therefore  MN  +  RS  =  MS. 
Illustration  II. 

Case  1.    To  add  the  line  RS  to  the  line  MN,  when  MN  is  longer 
than  RS. 


S^  ^  R  M ± ^ 

As  before,  starting  at  the  initial  point,  draw  the  hne  MN;  add  the 
line  RS  by  placing  the  point  R  on  the  point  N,  taking  care  to  keep  the 
lines  in  their  original  directions.     (See  note  on  p.  57.) 

M ± ^ 


S  -  i2 

Therefore  MN  +  RS  =  MS.    MS  is  a  positive  line. 

Case  2.    To  add  line  RS  to  line  MN,  when  MN  is  shorter  than  RS. 


As  before,  start  at  the  initial  point,  and  draw  line  MN;  add  the^line 


VI.  §  331  ADDITION  57 

RS  by  placing  point  R  on  point  N,  care  being  taken  to  keep  the  lines 
in  their  original  directions. 


Therefore  MN  +  RS  =  MS.    MS  is  a  negative  line. 

IlliLstration  III. 

Case  1.     To  add  line  RS  to  line  MN,  when  MN  is  longer  than  RS, 


^         N - M 


Adding  according  to  the  instructions  given  before,  we  have 
N  -  M 

Therefore  MN  +  RS  =  MS.    MS  in  this  case  is  negative. 

Case  2.     To  add  Hne  RS  to  hne  MN,  when  MN  is  shorter  than  RS, 

B 4-      £         N ~  Jg 


Adding  as  before, 


N^  — M 


R 

Therefore  MN  +  RS  =  MS.    MS  ia  &  positive  line. 

Illustration  IV. 

To  add  line  RS  to  line  MN. 

^^  —  R         i^  —  M 


Adding  as  before, 


N^ -- M 


S^  R 

Therefore  MN  +  RS  =  MS.    MS  is  a  negative  line. 

Note.  The  student  must  keep  in  mind  that  in  Illustrations  11  and 
m,  the  lines  MN  and  RS  coincide  —  he  one  on  the  other  —  and  that 
in  the  picture  one  is  drawn  a  httle  below  the  other  to  show  the  directions. 


58  ALGEBRA  — FIRST  COURSE  ivi.§34 

34.  Geometric  Addition  of  Numbers.  We  are  now  ready 
to  illustrate  what  we  mean  by  adding  two  numbers  together, 
whatever  may  be  their  sign.     Let  the  numbers  be  a  and  b. 

First  make  a  number  scale.  Starting  from  zero  on  this  scale, 
lay  off  a  line  segment  to  represent  the  number  a  in  amount 
and  direction;  using  the  point  that  you  have  now  reached  as 
an  initial  point,  lay  off  a  line  segment  to  represent  the  number 
b  in  amount  and  direction;  then  the  number  represented  by  the 
line  from  the  origin  or  zero  point  to  the  point  last  reached  is 
the  sum  a  -\-  b. 

As  shown  above  line  segments  can  be  added  when  both 
are  positive,  both  negative,  one  positive  and  one  negative; 
so  the  numbers  which  they  represent  may  be  added,  when 
both  are  positive,  both  negative,  one  positive  and  the  other 
negative. 

If  a  and  b  are  both  positive,  the  addition  corresponds  to 
the  addition  of  line  segments  as  shown  in  Illustration  I. 

Suppose  a  is  3  and  6  is  4;  we  have 

-10             1             2             3             U             5             6            7            8 
1 1 \ I I I I I L_ 


3  +  4  =  7. 

Suppose  a  is  7  and  6  is  —  3;  this  corresponds  to  Illustra- 
tion II,  Case  1,  in  the  addition  of  line  segments. 

-10  123U5G78 


^ 


7 +(-3)  =4. 

Suppose  a  is  —  7  and  6  is  3;   this  corresponds  to  Illustra- 
tion III,  Case  1,  in  the  addition  of  line  segments. 

—8          -7         -6         -5,         -U         -5         -^         -1            0            I 
^ I t  ^  I I I I I L 

-7  +  3  =-4. 


VI,  §  34J 


ADDITION 


59 


In  general,  if  the  two  numbers  are  a  and  6,  a  being  positive 
and  h  negative  we  have  the  two  figures  below.  The  first  is 
for  the  case  where  the  number  of  units  in  h  is  less  than  the 
number  of  units  in  a,  and  the  second  is  for  the  case  where 
the  number  of  imits  in  6  is  greater  than  the  number  of  units 
in  a. 


-3 


a-^h 


>!^ 


•^a 


L 


-I 


^ 


c-j-6 


■»ia 


In  these  figures,  as  in  the  case  of  preceding  figures,  the 
fact  that  h  stands  for  a  negative  number  is  shown  by  the 
arrow.  The  sum  of  a  and  6,  a  +  6,  is  positive  in  the  first 
figure  and  negative  in  the  second. 

We  may  now  state  the  following  rule: 

If  a  is  positive  and  h  is  negative  then 


the  sum  a -\- b  is 


(  positive  if\h\  is  less  than  a; 


i  negative  if\b\  is  greater  than  a. 
If  both  numbers  are  negative,  the  two  directed  lines  will 
both  extend  in  the  negative  direction;   hence  the  sum  of  two 
negative  numbers  is  a  negative  number  whose  absolute  value 
equals  the  sum  of  the  absolute  values  of  the  two  numbers. 


"6        -5        -4*        -s        -2 

I       . I ! I ^_L_ 


W 


_K- 


a+6 

This  corresponds  to  Illustration  IV  in  the  addition  of  line 
segments. 

Estimate  the  values  of  a  and  b  in  the  above  figures. 


60  ALGEBRA  —  FIRST  COURSE  ivi,§35 

Addition  of  more  than  two  numbers. 

Example.    Add  4  +  (-  6)  +  8. 

To  do  this  we  first  add  two  of  the  numbers;  then  to  their  sum  we 
add  the  third. 

This  is  shown  in  the  diagram  below;  4  +  (-  6)  leads  to  -  2;  add- 
ing +  8  to  -  2  leads  to  +  6. 


Algebraically  expressed  this  is 

4  + (-6) +8  =  6. 

35.  Exercises  and  Problems. 

Add  the  following  geometrically,  representing  the  num- 
bers by  lengths  of  lines.  The  best  results  will  be  obtained 
from  the  use  of  squared  paper.  Care  should  be  taken  in  each 
example,  especially  when  fractions  occur,  to  select  the  unit 
which  may  be  most  conveniently  used  in  the  example. 
Also  have  care  that  the  arrowheads  are  always  placed  to 
indicate  the  direction  of  the  line.  Write  the  algebraic  solu- 
tion in  each  case. 


1. 

6  +  3.                                    6.    140+  (-300). 

2. 

-12  +  2.                              6.   42 +(-8) +  (-3). 

3. 

2 +(-3) +  14.                   7.    -52 +  40 +(-3). 

4. 

i+(-i).                            8.    -15  +  4+ (-10). 

9.    -3  + (-10) +  (-2) +32. 

10.   .7  +  (-  .3)  +  (-  .05)  +  1.4. 

11.   3.7  +  5.7  +  (- 4)  +  7. 

12.    i+(-l)  +  (-i). 

13.  -^  +  (-1) +  !  +  (-§). 

In  each  of  the  following  problems  give  graphic  represen- 
tation, algebraic  expression  and  result. 

14.  A  boy  is  pulling  a  small  wagon  along  with  a  force  of 
5  lbs. ;  another  boy  comes  up  behind  and  pulls  back  with  a 
force  of  20  lbs.  In  what  direction  and  with  what  force  is  the 
wagon  pulled? 


VI,  §  35] 


PROBLEMS  61 


Solviion:  Let  the  line  —  represent  a  pound  force.    Then  the  following 
wiU  represent  the  forces  and  their  sum. 

-15  -10  -5  0  5  10 

<     <     I 1 I I I I I I 1 I I 1 I I I l—J I I I I I I I — I — L. 


The  algebraic  expression  is 

51b. +  (-20  lb.)  =-15  lb. 
The  negative  sign  shows  that  the  resulting  pull  is  backward. 

16.  A  man  earned  $5  on  Monday  and  spent  $3;  on 
Tuesday  he  earned  $2  and  spent  $6;  on  Wednesday  he 
earned  $7  and  spent  none;  on  Thursday  he  earned  $10  and 
spent  $4;  on  Friday  he  earned  $7  and  spent  $7.  How  much 
had  he  at  the  end  of  each  day?  What  does  a  negative  an- 
swer mean? 

16.  When  the  mercury  in  a  Fahrenheit  thermometer  reg- 
istered 73  degrees,  the  bulb  of  the  thermometer  was  in  a 
bottle  of  ether.  It  was  taken  out  quickly.  As  the  ether  on 
the  bulb  evaporated,  the  mercury  fell  2  degrees.  The  ther- 
mometer was  then  placed  in  hot  water  and  the  mercury 
rose  56  degrees.  What  temperature  did  the  thermometer 
then  register? 

17.  A  thermometer  (at  25  degrees)  was  placed  in  finely 
crushed  ice,  and  the  mercury  fell  24.6  degrees.  The  ther- 
mometer was  then  placed  in  a  mixture  of  salt  and  ice,  and 
the  mercury  fell  19  degrees  more.  When  it  was  finally 
placed  in  hot  water,  the  mercury  rose  108.2  degrees.  What 
was  the  temperature  of  the  hot  water? 

18.  On  consecutive  days  one  winter  the  following  was 
noticed:  On  the  morning  of  the  first  day  the  thermometer 
registered  zero;  by  the  next  morning  the  mercury  had  risen 
23  degrees;  by  the  following  morning  it  had  fallen  9  degrees; 
by  the  following  morning  it  had  fallen  25  degrees  more;  and 
by  the  next  morning  it  had  risen  7  degrees.  What  did  the 
thermometer  then  register? 


62  ALGEBRA  —  FIRST  COURSE  ivi.  §  35 

19.  A  boy  wished  to  have  the  water  in  his  beaker  a  cer- 
tain temperature.  He  tested  the  hydrant  water  and  found 
that  it  registered  34  degrees.  He  poured  in  hot  water  and 
raised  it  23  degrees;  he  next  poured  in  cold  water  and 
lowered  it  12  degrees;  then  hot  water  and  raised  it  3  degrees; 
then  cold  water,  and  lowered  it  7  degrees;  then  cold  water, 
and  again  lowered  it  2  degrees,  when  he  found  that  it  was 
the  temperature  that  he  wished  it  to  be.  What  was  the  final 
temperature? 

20.  Make  a  few  experiments  using  the  thermometer,  not- 
ing changes  of  temperature.  Write  problems  from  these 
observations. 

The  rule  for  adding  angles  is  similar  to  that  for  adding 
lines.  Draw  the  initial  arm  of  the  first  angle.  It  is  custom- 
ary to  have  this  extend  from  left  to  right.  Using  your  pro- 
tractor count  a  angular  units,  and  draw  the  final  arm  of 
your  first  angle;  using  this  as  the  initial  arm  of  the  second 
angle  count  h  angular  units  and  draw  the  final  arm  of  the 
second  angle.  The  sum  is  the  angle  with  Its  initial  arm  the 
same  as  the  initial  arm  of  the  first  angle  and  its  final  arm 
the  last  line  drawn. 

In  making  the  drawing  lay  off  a  positive  angle  so  that 
the  turning  from  the  initial  arm  toward  the  final  arm  is 
opposite  to  the  turning  of  the  hands  of  a  clock,  that  is, 
counterclockwise;  lay  off  negative  angles  in  the  clockwise 
direction. 

Illustration.  Add  an  angle  of 
—  44  degrees  to  an  angle  of  25 
degrees. 

Reading  from  the  figure  we 
have: 

Z  CAD-\-  ZDAB  =  ZCAB. 

The  algebraic  expression  is: 

25*'  +  (-44°) 19^ 


VI.  §  36]  SUBTRACTION  63 

Add  the  following;  give  drawing  and  algebraic  equation. 
Use  a  protractor,  so  as  to  have  the  work  as  accurate  as  possible. 

21.  25° +(-34°). 

22.  -  13°  +  56°  +  (-  12°). 

23.  43°  +  (-  156°)  +  73°. 

24.  43°  +  (-  94°)  +  141°  +  40°. 

26.  J  radian  +  f  radian  +  (—  i)  radian. 

26.  —  I  radian  +  i  radian  -\-  {—  i)  radian. 

27.  I  IT  radian  +  (—  f  tt)  radian. 

28.  IJ  X  radian  +  (—  27r)  radian  +  (—  Jtt)  radian. 

36.  Subtraction.  In  the  preceding  exercises  about  forces 
we  assumed  that  we  knew  the  number  of  pounds  and  the 
direction  of  each  force,  and  we  had  to  find  the  result  of  their 
combined  action;  that  is,  the  number  of  pounds  and  the 
direction  of  the  force  that  results  from  all  the  forces  com- 
bined in  pulling  and  pushing  the  body  in  one  direction  or 
the  opposite.  It  may  be  just  as  necessary,  in  practical 
work,  to  be  able  to  tell  what  force  must  be  combined  with 
another  to  secure  a  desired  result.  For  example,  what  force 
must  be  combined  with  a  downward  force  of  2  pounds  to 
get  a  resultant  upward  force  of  5  pounds?  Plainly  this  is 
the  inverse  of  the  work  that  we  have  been  doing.  So  we 
determine  the  answer  to  this  problem  by  guessing  the  amount 
and  then  adding  to  test  whether  we  have  guessed  correctly 
or  not.  After  a  while  we  become  able  to  guess  and  test  the 
result  very  rapidly. 

Definition  of  Subtraction.  The  process  of  determining  the 
amount  which  must  he  added  to  one  number  to  produce  another 
is  called  subtraction. 

What  we  have  said  about  forces  will  apply  to  all  the 
other  subjects  that  we  treated  in  addition. 

Exercises.  Give  answers  to  the  following  by  guessing  the 
answer  and  then  adding  to  test  the  correctness  of  the 
guess. 


64  ALGEBRA  —  FIRST  COURSE  [Vi,§36 

1.  A  mass  of  iron  filings  lying  on  the  scales  weighs  23 
grams.  What  mass  of  filings  must  be  added  in  order  to  make 
the  mass  weigh  37  grams? 

2.  The  thermometer  registers  69  degrees.  How  many 
degrees  must  be  added  in  order  to  make  it  register  40  degrees? 

3.  The  pressure  of  the  atmosphere  at  one  time  sustained 
in  a  barometer  a  column  of  mercury  73.1  centimeters  high, 
and  at  another  time  74.7  centimeters  high.  How  many 
centimeters  have  been  added  by  the  change  of  pressure? 

4.  At  what  rate  can  a  man  row  in  still  water  if  the  rate  of 
a  stream  is  3  miles  an  hour,  and  if  he  can  row  at  the  rate  of 
3  miles  an  hour  upstream? 

5.  A  boatman  rowing  upstream  finds  that  at  one  point 
he  is  not  moving.  If  his  rate  of  rowing  is  5  miles  an  hour, 
what  is  the  rate  of  the  stream  at  that  point? 

Subtraction  of  Lines. 

lUtistration.    Subtract  the  line  RS  from  the  line  PQ. 


^     p s 


The  question  is  —  What  length  of  line  must  be  added  to  the  line  RS 
to  get  the  line  PQ? 

Examining  your  graphic  representation  of  the  addition  of  two 
directed  lines,  you  will  observe  that  the  line  you  added  to  and  the  line 
which  is  the  sum  of  the  two  lines  both  have  the  same  initial  point. 
The  line  added  on  extends  from  the  final  point  of  the  line  added  to,  to 
the  final  point  of  the  sum.  From  this  we  see  that,  when  we  are 
given  a  line  and  wish  to  determine  the  length  and  direction  of  a  line 
that  must  be  added  to  it  in  order  to  obtain  another  line,  which  is  in 
reaUty  the  sum  of  the  given  line  and  the  one  to  be  found,  we  place  the 
two  given  Unes  with  their  initial  points  together,  taking  care  that 
each  keeps  its  original  direction.  Draw  a  line  from  the  final  point  of 
the  line  to  be  added  to,  to  the  final  point  of  the  Une  which  is  the  sum. 
This  is  the  line  we  seek. 

That  is  place  Une  RS  and  line  PQ  with  the  point  R  on  the  point  P 
taking  care  that  each  keeps  its  original  direction.  Draw  a  line  from 
point  S  to  point  Q,    The  line  SQ  is  the  line  which  we  are  seeking; 


VI.  §  36]  SUBTRACTION  65 

that  is,  it  is  the  line  which  must  be  added  to  the  line  RS  to  obtain  the 
line  PQ.     (See  note  on  p.  57.) 

Line  PQ  —  line  RS  =  line  SQ.    SQ  is  a  negative  line. 
That  is,  SQ  must  be  added  to  RS  to  give  PQ. 

Exercises.  Subtract  the  first  of  the  following  lines  from  the 
second,  observing  carefully  the  above  remarks. 

B< A    D S 

M f-N  i?-< K 

Q^ p      S-* R 

Subtraction  of  Numbers. 

Illustration.  If  you  have  a  line  3  units  long,  what  length  of  line  must 
you  add  to  it  in  order  to  have  a  Une  —  2  units  long? 

This  does  not  differ  from  the  preceding  work  excepting  that  we  use 
the  scale  of  measurement  in  order  to  do  the  work  more  quickly.  We 
follow  the  same  rule  as  above. 

-3                  -2                  ~l                    0                     1                     2  3 

^J 1 \ ! 1 I 


Thus  we  see  that  we  must  add  a  line  negative  5  units  long  to  a  line 
3  units  long  in  order  to  have  a  line  negative  2  units  long. 
The  algebraic  expression  for  this  is 

(-2) -(+3)  =-5. 

This  is  read,  "Negative  2  minus  3  equals  negative  5,"  or  "3  sub- 
tracted from  negative  2  gives  negative  5,"  and  means  that  which  must 
be  added  to  3  to  get  negative  2. 

The  check  to  this  is:  3  +  ( -  5)  =  -2. 

Exercises.  In  the  following,  give  graphic  representation, 
algebraic  expression,  result,  and  check. 

1.  What  length  of  line  must  be  added  to  a  line  negative 
3  units  long  in  order  to  have  a  line  positive  8  units  long? 

2.  Subtract  a  line  positive  28  cm.  long  from  a  line  positive 
23  cm.  long.    We  see  readily  that  this  would  have  little 


66  ALGEBRA  —  FIRST  COURSE  [VI.  §36 

meaning  if  we  did  not  hold  to  our  definition  of  the  word 
subtract.  The  question  in  reality  is  —  What  must  be  added 
to  the  line  28  cm.  in  length  (taking  into  consideration  both 
distance  and  direction)  to  have  a  line  23  cm.  long? 

3.  What  must  be  added  to  a  line  11  units  long  to  have  a 
line  2  units  long? 

4.  What  must  be  added  to  a  line  negative  17  units  in 
length  to  have  a  line  negative  2i  units  in  length? 

5.  What  must  be  added  to  a  line  54  mm.  long  to  have  a 
line  0  mm.  long? 

6.  What  must  be  added  to  a  line  —  7f  cm.  long  to  have  a 
line  —  2f  cm.  long? 

7.  What  must  be  added  to  a  line  4^  cm.  long  to  have  a 
line  2f  cm.  long? 

8.  Repeat  the  idea  of  the  questions  asked  in  Exercises  3 
to  7  using  the  word  subtract. 

9.  If  a  boy  is  pulling  a  block  with  a  force  of  6  lb.,  with 
how  much  force  and  in  what  direction  must  another  boy 
pull,  if  the  net  result  is  to  be  a  pull  of  5  lb.  in  the  opposite 
direction?     How  else  may  this  question  be  asked? 

A  graphic  representation  may  be  made  of  this  exercise  in 
the  same  way  that  we  represented  the  exercises  in  the  addi- 
tion of  forces.  That  is,  we  let  a  certain  length  of  line  repre- 
sent the  unit  of  force,  and  by  taking  multiple  lengths  of 
this  line  or  fractional  parts  of  it,  we  can  make  a  line  repre- 
sentation of  the  forces,  which  lines  may  be  subtracted  in 
the  same  manner  as  is  given  above. 

Let  one  unit  of  length  represent  one  pound  force.  Then 
the  graphic  representation  of  the  subtraction  is  as  follows. 

-5  Q  6 

'       ' I 1 I 


-11 
The  algebraic  expression  for  this  is 

(-5)-  (+6)  =-11     or,     -5 -6  =-11, 


VI.  §  36]  SUBTRACTION  67 

Such  a  graphic  representation  may  be  made  of  any  meas- 
urable quantities. 

Make  graphic  representation  of  the  following  exercises; 
give  algebraic  expression  and  result  as  in  preceding  exer- 
cises. 

10.  If  a  man  earns  ten  dollars,  how  much  must  he  spend  to 
be  in  debt  four  dollars?  Since  spending  is  the  opposite  of 
earning,  this  may  be  asked  —  How  much  must  a  m^n  add 
to  ten  dollars  in  order  to  have  negative  four  dollars? 

11.  If  wheat  is  $1.05  a  bushel  on  one  day,  and  $.93  the 
next,  what  has  been  added  to  the  price  the  first  day  to  get 
that  of  the  second? 

12.  If  certain  shares  of  stock  sell  at  5  cents  above  par  one 
day,  and  2  cents  below  the  next,  what  has  been  added  to 
the  price  of  the  first  day  to  get  that  of  the  next? 

13.  A  strip  of  iron  when  warm  extended  past  a  certain 
notch  3  cm.;  when  cooler,  the  end  touched  the  notch.  If 
we  call  the  effects  of  expansion  positive,  how  much  was 
added  to  the  length  of  the  strip  by  contraction?  If  when 
cold  the  strip  lacked  2  cm.  of  reaching  to  the  notch,  how 
much  was  added  to  the  original  length  by  contraction? 

14.  An  iron  nail  lying  in  a  pan  on  the  scales  weighs  10 
grams.  When  a  magnet  is  held  over  it,  the  scales  register 
2  grams.  If  we  regard  an  upward  pull  as  positive,  what  is 
the  pull  of  the  magnet  upon  the  nail? 

Write  problems  leading  to  some  of  the  following  algebraic 
expressions;  give  graphic  representation  and  answers  for 
all. 


15. 

9-2. 

22. 

Subtract  8  from  7. 

16. 

-  22  -  8. 

23. 

Subtract  3J  from  1|. 

17. 

-  22  -  (  - 

9). 

24. 

Subtract  3.6  from  .7. 

18. 

0  -  (-  3). 

26. 

Subtract  20.8  from  -  .9. 

19. 

0-3. 

26. 

Subtract  -  .08  from  .25. 

20. 

0  -  (-  15). 

27. 

Subtract  —  1.7  from  —  .7. 

21. 

-7-0. 

28. 

Subtract  —  f  from  —  f . 

68  ALGEBRA  —  FIRST  COURSE  IVI.  §  36 

29.  From  an  angle  of  30  degrees  subtract  an  angle  of  45 
degrees. 

Illustration:  The  question  is — 
What  must  be  added  to  an  angle 
of  45  degrees  to  get  an  angle  of ^30 
degrees? 

The  symbolic  expressions  are 

ZBAC  -  Z  BAD  =  ZDAC. 
30°  -  45°  =  -  15°. 
Check:  ZBAD+  ZDAC=  ABAC. 

45°  +  (- 15°)  =30°. 

For  the  exercises  below  make  geometric  picture;  give  alge- 
braic expression  and  result. 

30.  —  I TT  radians  —  |  tt  radians. 

31.  i  TT  radians  —  x  radians. 

32.  27°  -  125°. 

33.  -  45°  -  (-  60°). 

34.  —  J  TT  radians  —  ( —  f  tt)  radians. 

35.  What  angle  must  be  added  to  10  degrees  to  get  a 
right  angle? 

36.  What  angle  must  be  added  to  37  degrees  to  get  a 
right  angle? 

37.  What  angle  must  be  added  to  ix  radians  to  get  a 
right  angle? 

Two  angles  whose  sum  is  a  right  angle  are  called  comple- 
mentary angles.  Either  angle  is  said  to  be  the  complement 
of  the  other.  In  the  last  three  exercises  you  have  found 
the  complements  of  the  angles  given. 

38.  What  is  the  complement  of  80  degrees?  of  53  de- 
grees? of  —  i  X  radians?  of  —  tt  radians?  of  175  degrees? 
of  0  degrees?  of  90  degrees? 

39.  What  angle  must  be  added  to  10  degrees  to  get  a 
straight  angle? 

40.  What  angle  must  be  added  to  fx  radians  to  get  a 
straight  angle? 


VI.  §  361  SUBTRACTION  69 

Two  angles  whose  sum  is  a  straight  angle  are  called  sup- 
plementary angles.  Either  angle  is  said  to  be  the  supple- 
ment of  the  other.  In  the  last  two  exercises  you  have  found 
the  supplements  of  the  angles  given. 

What  are  the  supplements  of  the  angles  given  in  Exercise 
38? 

Exercises  for  Quick  Oral  DriU.  The  teacher  may  read  the 
following  or  similar  exercises,  and  the  student  should  be 
able  to  give  the  answer  as  soon  as  the  exercise  is  read.  Read 
the  subtraction  exercises  in  three  different  ways. 

1.  2  -  (-  3).  4.       -  2  -  (-  6). 

2.  5  -  (-  4).  5.    8  -  9. 
3.-5-4.  6.   3  +  r  +  (,-  4). 

7.   5 +  (-2) +  (-3)  4-2  +  7. 

8.  3  -  (-  7).  13.   3  H-  6  +  (-  15)  +  9. 

9.  -  3  -  (-  7).  14.    0  -  12. 

10.  5-8.  16.   0  -  (-  12). 

11.  -  15  -  (-  3).  16.   3  -  21. 

12.  4  -  12.  17.   21  -  (-  16). 

18.  -  13  + (-2)  +  10  +  3. 

19.  -5  +  7+ (-8) +7  +  8. 

20.  -  2  +  (-  3)  +5.  22.    -  2  -  0. 

21.  0  -  (-  4).  23.   0-2. 

24.    -4  +  5+ (-5)  + (-3). 
25.    I  -  1  -  J  26.    f  -  2f . 

27.  -§+|  +  (-f)  +  |. 

28.  tV  +  (-t\)  +  (-^)+tV 

29.  -  i^j  +  (-  ^%)  +  B  +  (-  ^y. 

30.  U  -  (-  H).  34.  120  +  (-  125). 

31.  120  -  125.  35.  -  120  -  125. 

32.  il  +  (-  tV).  36.  -  120  +  125. 

33.  \l  -  tV-  37.  -  120  -  (-  125). 

38.    -15  +  17+(-l)  +  (-3). 

39.  22  +  (-  27)  +  5  +  3.       42.   5  -  (-  6). 

40.  -5  + (-6).  43.   5  +  6. 

41.  -5-6.  44.   37 +(-41) +  5. 


70  ALGEBRA  — FIRST  COURSE  IVI.§37 

37.   Summary. 

Addition  of  directed  lines.  To  add  a  given  directed  line 
to  another,  place  the  initial  point  of  the  line  you  are  adding, 
on  the  final  point  of  the  line  you  are  adding  to,  having  care 
that  both  lines  keep  their  original  direction.  The  distance 
and  direction  of  the  line  extending  from  the  initial  point  of 
the  line  added  to,  to  the  final  point  of  the  line  added  on,  is 
the  length  and  direction  of  the  sum. 

Comparison  of  measurable  quantities  may  be  expressed  by 
means  of  lines  as  well  as  by  numerals  and  letters. 

To  express  the  addition  of  quantities  by  the  addition  of 
lines,  assume  a  unit  length  of  line  to  express  a  unit  of  quan- 
tity. By  taking  multiples  or  fractional  parts  of  this  unit, 
draw  lines  to  express  the  quantities  to  be  added.  Add 
these  lines  by  the  above  rule. 

Subtraction  of  directed  lines.  Subtraction  is  the  inverse  of 
addition.  It  consists  in  finding  the  quantity  which  when 
added  to  one  stated  quantity  will  give  another  stated 
quantity. 

To  subtract  one  directed  line  from  another  directed  line, 
place  the  two  lines  with  their  initial  points  together  having 
care  that  the  lines  keep  their  original  direction.  The  length 
and  the  direction  of  the  line  extending  from  the  final  point 
of  the  line  to  be  added  to,  to  the  final  point  of  the  sum,  is 
the  line  required. 

To  express  subtraction  of  quantities  by  subtraction  of 
lines,  express  their  values  by  means  of  lines  as  instructed  in 
addition,  then  subtract  these  lines  by  the  rule  just  given. 

The  terms  minuend  and  subtrahend  are  used  in  the  same 
sense  as  in  arithmetic. 


CHAPTER  VII 
MULTIPLICATION   AND   DIVISION 

38.  Meaning  of  Multiplication.  The  terms  multiplicand 
and  multiplier  are  used  in  the  same  sense  as  in  arithmetic. 

The  subject  of  multiplication  adds  nothing  new  to  the 
work  we  have  been  doing.  So  far,  the  numbers  that  we 
have  been  dealing  with  have  told  us  what  to  do  with  a 
unit  of  measure  if  we  wished  to  express  the  measurement  of 
quantity.  For  example,  when  we  say  that  a  line  is  negative 
5  cm.  in  length,  we  mean  to  take  the  unit,  turn  it  over,  and 
use  it  five  times  to  get  the  length  of  line  called  for.  Now 
instead  of  a  single  unit  we  shall  use  multiples  of  or  fractional 
parts  of  the  unit,  using  them  in  either  the  same  direction  as 
the  unit  or  in  the  reverse  direction;  these  numbers  are 
multiplicands,  and  the  multiplier  tells  us  what  to  do  with  them 
just  as  the  number  in  the  example  above  told  us  what  to  do 
with  the  unit  to  get  it.     In  other  words. 

To  multiply  a  second  number  by  a  first,  we  are  to  do  to  the 
second  number  the  same  thing  that  we  did  to  the  unit  to  get  the 
first. 

39.  Graphic  Illustrations  of  Multiplication. 

Illustration  I.  A  line  three  units  long  lying  in  the  positive  direc- 
tion must  be  used  five  times  to  measure  a  line;  what  is  the  length  of 
the  Une? 

Unit  Measuring  Line 

0  S  6  9  IS  15 

•^ 1 1 1 1 1 1 1 1 1 1 1 1 1 1 >4 

0  I'S  2-3  3-3  lc$  5-3 

Our  hne  is  15  units  long. 
The  algebraic  expression  of  this  operation  is 
5  .  3  =  15. 
71 


72  ALGEBRA  —  FIRST  COURSE  [Vil.§39 

It  i3  read  5  times  3  equals  15. 

Did  we  use  the  3,  in  this  case,  in  the  same  way  that  we  would  have 
used  the  unit  had  we  been  asked  to  lay  off  a  line  5  units  long?  Does 
our  definition  of  multiplication  hold  in  this  case? 

Illustration  II.  A  line  3  units  long  is  lying  in  the  positive  direction. 
Reverse  it  and  use  it  5  times  to  measure  a  hne.  What  length  of  line 
is  measured?  What  is  its  direction?  What  sign  is  used  to  indicate 
"reverse"  direction? 

The  geometric  representation  of  this  is 

Unit  Measuring  Line 


'15  -Jf  -9  -6  -3 

I I I I 1 \ 1 1 t  ,.       I 

-Ji-2  -S-2  -2-3  "1-3 


•IT" 


The  algebraic  expression  of  this  is 

-  5  .  3  =  -  15. 

It  is  read  negative  5  times  3  equals  negative  15. 

In  this  case  did  you  do  the  same  thing  to  3  that  you  would  have  had 
to  do  to  the  unit  to  get  negative  5? 

Illustration  III.  A  line  3  units  long  is  lying  in  the  negative  direc- 
tion (opposite  direction  to  the  unit).  Keeping  it  in  this  direction  lay 
it  off  5  times  along  the  line  to  be  measured.  What  is  the  length  of 
line  measured?  What  is  its  direction?  What  is  the  sign  used  to  indi- 
cate "  keep  same  direction  "  ? 

The  geometric  representation  of  this  is 


p5ti< 


Measuring  Line 
\< 1 ' ^ 


-Jl  -9  -«  -» 


5(23)  U-S)  S(.-3)  Si-S)  3  (-3)  0 

The  algebraic  expression  of  this  is 

5.  (-3)  =-15. 

It  is  read  5  times  negative  3  equals  negative  15. 

Illustration  IV.  A  hne  3  units  long  is  lying  in  the  negative  direc- 
tion. Reverse  it,  and  lay  it  off  5  times  along  the  line  to  be  measured. 
What  is  the  length  of  Une  measured?  What  is  its  direction?  What 
sign  is  used  to  denote  reverse  direction? 


VII,  §  39]  MULTIPLICATION  73 

The  geometric  representation  of  thia  is 


Unit                      Measuring  Line 
H  *< ' ' ' 


0  3  6  9  IS  15 

I I I I I I I I I I I I I I I       ^1 

0  -l{-3)  -2  (-3)  -3  (-2)  -Ui-S) 


-5tr3) 


The  algebraic  expression  of  this  is 

^5.  (-3)  =  15. 
It  is  read,  negative  5  times  negative  3  equals  15. 

Exercises. 

1.  Repeat  Illustrations  I,  II,  III,  IV,  using  a  line  f  of  a 
unit  in  length  as  the  measuring  line.  Give  geometric  rep- 
resentation, and  algebraic  expression. 

2.  Give  geometric  representation  and  result  of  the  fol- 
lowing: 

f.4;     f.(-4);     -f.4;     -  f  •  (- 4). 

3.  If  a  piece  of  iron  weighs  25  5-gram  weights,  and  a 
piece  of  wood  weighs  7  5-gram  weights,  and  a  piece  of  stone 
weighs  10  5-gram  weights,  how  many  5-gram  weights  do 
the  pieces  together  weigh?  How  many  grams  do  they 
weigh? 

4.  If  three  boys,  capable  of  pulling  an  average  of  75 
pounds  apiece,  are  pulling  a  sled,  and  seven  boys,  with  the 
same  average  pulling  capacity,  come  up  and  pull  in  the  oppo- 
site direction,  in  what  direction  will  the  sled  be  pulled,  and 
with  how  many  times  75  pounds  force?  With  how  many 
pounds  force? 

6.  If  you  go  3  times  3J  units  forward,  then  6  times  3| 
units  backward,  how  many  3J  units  are  you  from  the  start- 
ing point? 

6.  If  you  go  3  times  —  2  units  forward,  then  6  times  —  2 
units  backward,  how  many  times  —  2  units  are  you  from 
the  starting  point?     How  many  units? 


74  ALGEBRA  —  FIRST  COURSE  ivii.§40 

7.  If  you  go  3  times  —  3f  units  forward  and  6  times  —  3J 
units  backward,  how  many  times  —  3^  units  are  you  from 
your  starting  point? 

8.  If  you  go  3  times  n  imits  forward  and  6  times  n  units 
backward,  how  many  times  n  units  are  you  from  your 
starting  point? 

9.  A  load  is  being  pulled  in  a  certain  direction  with  a 
force  of  3  100-pounds.  What  force  must  be  added  in  order 
that  it  may  be  pulled  in  the  opposite  direction  with  a  force 
of  17  100-pounds? 

10.  Two  forces  are  acting  in  opposite  directions  with  a 
resultant  force  of  6.4  5-pounds.  If  one  of  the  forces  is 
known  to  be  151.2  5-pounds,  what  is  the  amount  and  direc- 
tion of  the  other  force? 

11.  If  an  engine  has  gone  17  m-meters  from  the  station, 
how  far  and  in  what  direction  must  it  go  to  be  —  3  m-meters 
from  the  station. 

12.  If  the  temperature  rises  4  d  degrees,  then  falls  7  d 
degrees,  what  is  the  total  change  in  temperature? 

40.  Multiplication  of  General  Numbers.  In  the  preced- 
ing exercises  we  brought  out  the  following  facts: 

If  the  sign  of  the  multiplier  is  positive,  the  sign  of  the  product 
is  the  same  as  the  sign  of  the  multiplicand. 

If  the  sign  of  the  multiplier  is  negative,  the  sign  of  the  prod- 
uct is  opposite  to  the  sign  of  the  multiplicand. 

In  other  words: 

A  positive  number  times  a  positive  number  gives  a  posi- 
tive nmnber. 

A  positive  number  times  a  negative  number  gives  a  nega- 
tive number. 

A  negative  number  tunes  a  positive  number  gives  a 
negative  number. 

A  negative  number  times  a  negative  number  gives  a 
positive  number. 

Would  these  facts  have  been  brought  out  had  we  used 


vri,  §  40] 


MULTIPLICATION  75 


other  numbers  in  our  exercises?  Yes,  you  will  say,  any 
numbers  might  have  been  used.  Let  us  then  use  the  letter 
m  to  stand  for  the  multiplier,  and  the  letter  n  to  stand  for 
the  multiplicand.  Such  letters  may  then  be  called  general 
numbers,  since  they  may  be  used  to  stand  for  any  number 
we  choose. 

Our  last  four  rules,  in  algebraic  form,  will  then  be 

m  *  n  =  fnn. 

m  •  (—  n)  =  —  mn. 
—  m*  n  =  —  mn. 

—  m  '  {—  n)  =  mn. 
Exercises. 

1.   The  distance  from  floor  to  floor  in  an  office  building  ia 
on  the  average  h  feet.     The  elevator  boy,  starting 
at  the  main  floor,  runs  the  cage  up  two  stories,  then      T 
down  three,  then  up  seven,  when  he  is  at  the  top        [_ 
floor.     How  many  feet  is  it  from  the  main  floor  to 
the  top  floor? 

The  geometric  solution  is  shown  in  the  adjacent 
diagram.     The  algebraic  expression  is: 

2A  +  (-3/i)  +  7A  =6A. 


6h- 
574- 


Sh 


Ih 


Furthermore,  we  could  bring  out  the  mathemati- 
cal thought  in  a  still  more  general  way  by  having 
the  example  read  thus: 

The  distance  from  floor  to  floor  in  an  office  build- 
ing is  on  the  average  h  feet.  The  elevator  boy, 
starting  from  the  main  floor,  runs  the  cage  up  a 
stories,  then  down  h  stories,  then  up  c  stories,  when 
he  is  at  the  top  floor.  How  many  feet  is  it  from 
the  main  floor  to  the  top  floor? 

The  algebraic  expression  for  the  second  state- 
ment, which  we  might  call  the  general  statement,  is 

ah  +{~bh)-\-ch=(a-b  +  c)  h. 


. 
' ' 


■ih^-^ 


76  ALGEBRA  —  FIRST  COURSE  ivii.§40 

When  we  express  numbers  by  means  of  letters,  we  have 
no  single  symbol  as  we  have  in  arithmetic  to  express  their 
sum,  so,  when  necessary  to  show  that  a  combination  of 
numbers  is  to  be  regarded  as  one  quantity  w^e  inclose  the 
numbers  in  parentheses  as  shown  above  (§8).  However, 
either  expression  in  the  second  statement  may  be  regarded 
as  the  answer  to  our  example. 

2.  A  man  made  it  a  point  to  have  his  average  daily  ex- 
penses the  same  as  the  rent  from  a  certain  piece  of  property 
which  he  owned.  After  t  days  during  the  month  of  October 
the  renter  of  the  property  moved  and  the  rent  stopped. 
Compute  the  net  amount  the  man  had  for  the  month,  if  the 
average  rent  per  day  was  r  dollars. 

The  work  of  writing  problems  leading  to  given  arithmetic 
or  algebraic  expressions  is  of  great  value.  It  is  the  best  way 
of  testing  a  student's  insight  into  the  meaning  of  algebraic 
symbols. 

When  a  problem  is  given  and  the  solution  is  called  for, 
the  student  arrives  at  the  answer  to  the  question  asked  by 
first  translating  the  English  statements  into  algebraic  sym- 
bol language  and  then  doing  the  work  called  for  by  these 
symbols. 

The  writing  of  a  problem  is  the  reverse  of  this  process. 
The  algebraic  expression  is  given  and  the  student  is  asked 
to  translate  its  meaning  into  English. 

The  correctness  of  the  problem  should  always  be  tested 
by  a  translation  back  into  the  algebraic  language;  that  is, 
by  the  solution  of  the  problem. 

Example  1.     Write  a  problem  leading  to  the  expression 
6  .  40  +  3  .  40. 

Solution :  Upon  examining  this  expression  we  see  that  there  are 
two  terms,  each  of  which  is  a  multiple  of  40.  So  we  must  write  a 
problem  in  which  40  is  repeated. 

Six  boys,  exerting  a  force  of  40  pounds  each,  are  pulling  on  a  load. 
Not  being  able  to  move  it,  they  call  to  their  assistance  3  other  boys 


VII.  §  411  DIVISION  77 

each  capable  of  exerting  a  force  of  40  pounds.  They  succeed  in  start- 
ing the  load  in  motion.  What  was  the  amount  of  force  necessary  to 
move  the  load? 

Check  :  Translating  into  algebraic  symbols: 

6  •  40  =  the  number  of  pounds  force  the  first  6  boys  exert; 
3  •  40  =  the  number  of  pounds  force  the  3  boys  exert; 

so  that 

6  •  40  +  3  •  40  =  the  number  of  pounds  necessary  to  move  the  load. 

6  •  40  +  3  •  40  =  9  •  40,  the  number  of  pounds  necessary  to  move  the 
load. 

Give  a  statement  of  a  problem  like  this,  using  general 
numbers. 

Example  2.     Write  a  problem  leading  to  the  expression 
6.  (- 15)  -6-25, 
or,  -  6  .  15  -  6  .  25. 

Solution  :  A  train  runs  due  north  for  6  hours  at  the  rate  of  25  miles 
an  hour.  At  what  rate  must  it  now  go  for  6  hours  so  as  to  arrive  at 
the  same  place  as  if  it  had  gone  south  from  the  original  starting  point 
for  6  hours  at  15  miles  an  hour? 

Here  we  regard  north  as  positive.  Check  this  statement.  Give  a 
similar  problem  involving  general  numbers. 

The  best  way  to  learn  to  write  these  problems  is  to  study 
carefully  the  expressions  obtained  from  problems  in  the 
preceding  lists.  Such  lists  are  given  not  only  for  the  solu- 
tion of  the  problems  but  to  show  how  such  problems  arise. 

Write  problems  for  some  of  the  following.  Give  graphic 
representation  and  result  for  all. 

3.  15.4  +  3.4.  9.    -6a  +  7a+(-3)a. 

4.  9.  (-6) -3.  (-6).  10.    -2(-c)+(-5)(-c). 
6.   7  .  (-  2/)  -  10  .  (-  y).         11.   m{-n)  +r(-  n). 

6.  3.f  +  (-7).f.  12.    -9(-p)-5(-p). 

7.  7wJ+ (-3)w?+(-4)iy.   13.   a(-6)  -  (-a)(-6). 

8.  -/+(-4)/.  14.   2/b-8/b. 

41.   Division. 

Division  0}  a  first  number  by  a  second  is  the  process  of  deter- 


78  ALGEBRA  —  FIRST  COURSE  IVII.§41 

mining  the  number  by  which  you  must  multiply  the  second 
number  to  get  the  first. 

It  is  the  inverse  operation  of  multiplication,  and  is  a 
purely  guessing  process.  You  guess  the  answer  and  mul- 
tiply to  see  if  your  guess  is  correct.  In  this  respect  it  does 
not  differ  from  the  work  in  arithmetic,  for  long  division  is 
nothing  more  than  a  series  of  guesses  and  multiplication  to 
test  the  correctness  of  the  guess. 

No  new  rules  need  be  given  for  division,  as  you  have  but 
to  observe  the  rules  for  multiplication  when  it  comes  to 
testing. 

Example.  You  are  to  divide  —  ahhy  b;  that  is,  you  want  to  know 
by  what  you  must  multiply  h  to  get  —  ab.  You  have  learned  that 
this  is  —  a.     So  —  ab  divided  by  b  equals  —  a. 

The  symbols  for  division  are  the  same  as  in  arithmetic.  So  sym- 
boUcally  we  write 

—  ab  -7-  b]     or     — r — 

Of  these  two  forms  the  latter  is  almost  always  used.     We  then  have 
—  ab 

Notice  that  ordinary  cancellation,  as  in  arithmetic,  can  be  used 
here  to  reduce  the  given  fraction.  The  only  difference  is  that  in  arith- 
metic you  used  only  positive  numbers. 

Problems.  In  these  problems  give  algebraic  expression  and 
result.     Multiply  to  check  your  answer. 

1.  A  chauffeur  finds  that  he  has  gone  7 1  miles  in  t  hours. 
What  is  his  speed  per  hour?  What  is  his  speed  per  hour  if 
he  goes  m  miles  in  t  hours? 

2.  A -boy  agrees  to  work  for  $1.50  per  day;  how  many 
days  must  he  work  to  earn  $6?  How  many  days  must  he 
work  to  earn  n  dollars,  if  he  earns  d  dollars  a  day? 

3.  A  cup  made  of  tin  in  the  form  of  a  cone  holds  v  grams 
of  water,  and  a  cup  in  the  form  of  a  cylinder  whose  height 
and  circumference  around  the  bottom  are  the  same  as  that 


VII,  §411  DIVISION  79 

of  the  cone,  holds  3  v  grams.     What  is  the  ratio  of  the 
volume  of  the  cylinder  to  that  of  the  cone? 

By  ratio  of  one  number  to  another  we  mean  the  number  by 
which  we  must  multiply  the  second  to  get  the  first. 

4.  One  cup  holds  mw  grams  and  another  holds  w  grams. 
What  is  the  ratio  of  the  capacity  of  the  two  cups? 

5.  If  a  boatman  rows  at  the  rate  of  v  miles  an  hour  in  a 
stream  that  flows  at  the  rate  of  s  miles  an  hour,  what  is  the 
ratio  of  his  rate  of  rowing  to  the  rate  of  the  stream? 

6.  If  an  automobile  goes  r  miles  an  hour  for  t  hours,  and 
another  automobile  goes  r  miles  an  hour  in  the  opposite 
direction  for  s  hours,  what  is  the  ratio  of  the  distance  the 
first  goes  to  the  distance  the  second  goes? 

7.  If  a  train  is  going  east  at  the  rate  of  r  miles  an  hour, 
what  is  the  ratio  of  its  velocity  to  that  of  a  train  going  west 
at  the  rate  of  r  miles  an  hour? 

8.  What  is  the  ratio  of  the  length  of  a  line  —  2  cm.  long 
to  the  length  of  a  line  7  cm.  long? 

9.  By  what  must  you  multiply  a  force  of  51  lbs.  in  order 
to  have  a  force  of  f  lb.?  Answer  this,  using  the  word 
ratio.     Check. 

10.  If  12  men  can  do  a  piece  of  work  in  a  day,  how  much 
can  3  men  do  in  a  day  if  the  average  amount  of  work  done 
by  each  is  the  same? 

11.  If  a  men  can  do  a  piece  of  work  in  m  days,  what  part 
of  the  work  will  1  man  do  in  1  day,  if  the  average  amount  of 
work  done  by  each  is  the  same? 

12.  By  what  must  a  line  —  5  units  long  be  multiplied  in 
order  to  have  a  line  —  12  units  long?  Answer  this,  using 
the  word  ratio. 

13.  Divide  a  line  —  1  cm.  long  by  a  line  r  cm.  long. 
Read  this,  using  the  word  ratio. 

14.  The  same  number  of  boys  were  trying  to  keep  a 
door  shut  as  the  number  that  were  trying  to  pull  it  open. 
The  amount  of  force  exerted  was  ma  —  mb  pounds.     What 


80  ALGEBRA  —  FIRST  COURSE  [Vii,§42 

is  the  number  of  boys  on  each  side,  if  the  average  number  of 
pounds  exerted  by  each  boy  on  one  side  is  a,  and  the  average 
number  of  pounds  exerted  by  each  on  the  other  side  is  6? 

SimpHfy  each  of  the  following  expressions  as  much  as 
possible.     State  problems  leading  to  some  of  them. 

15.    ^J.  20.    ^^.    -  26.    ^-  °>^'. 

8t  Uab^  -am 

"•    5a  ^^-      -b  ,  ^^-  xH-yy 

1,.   32»_^  22.   _6_.  '  27.   ^^^'(-'•^ 


8  71  —  ab  2^  pV 


18     4«  „,     (-h){~k)  ia{-b)' 

„    2«6  (-x)(-j/)(-z)       „.     (jymH-ny 

5  a  —xy  (f)2m%  — n) 

42.   Summary. 

Definition  of  multiplication.  To  multiply  a  second  num- 
ber by  a  first,  do  to  the  second  number  what  you  must  do 
to  the  unit  to  get  the  first. 

Law  of  signs  in  multiplication.  If  the  sign  of  the  multi- 
plier is  positive,  the  sign  of  the  product  is  the  same  as  that 
of  the  multiplicand. 

If  the  sign  of  the  multiplier  is  negative,  the  sign  of  the 
product  is  the  opposite  to  that  of  the  multiplicand. 

Literal  numbers  are  used  in  mathematics  when  we  wish 
to  bring  out  general  truths,  that  is,  truths  which  do  not 
depend  upon  the  exact  amounts,  but  upon  the  relation  of 
the  quantities  under  consideration. 

Division.    This  is  the  inverse  of  multiplication. 


CHAPTER  VIII 
ADDITION    AND    SUBTRACTION    OF    POLYNOMIALS 

43.  Definitions.  An  expression  consisting  of  several  parts 
connected  by  the  signs  +  or  —  is  called  a  polynomial. 

For  example, 

a  +  6,     2h  —  3k  +  1,    m-{-4:n  —  pq  +  Srs 

are  polynomials. 

Each  part  of  a  polynomial,  including  its  sign,  is  called  a 
term  of  the  polynomial. 

Thus  4  n  is  a  term  of  the  last  polynomial  written  above. 
Another  term  is  —  pq. 

An  expression  consisting  of  a  single  term  is  called  a 
monomial. 

A  polynomial  consisting  of  two  terms  is  called  a  binomial. 

A  polynomial  consisting  of  three  terms  is  called  a  tri- 
nomial. 

Two  terms  which  differ  only  by  a  numerical  factor,  as 
3  mn  and  —  5  mn,  are  called  similar  terms. 

44.  Addition  and  Subtraction  of  Polynomials. 

Example  1.  You  learned  in  your  arithmetic  and  reviewed  in  the 
chapter  on  measurement  the  fact  that  a  quantity  may  be  measured  by 
a  unit  and  subdivisions  of  that  unit. 

For  example  a  distance  is  measured  in  yards  and  feet.     Say  that  it 
measures  9  yards  and  2  feet.     This  may  be  written 
9  yards  +  2  feet. 

Another  distance  measures  3  yards  and  7  feet.  This  may  be 
written 

3  yards  +  7  feet. 

Then  the  sum  of  these  two  distances  is  12  yards  and  9  feet,  which 
may  be  written 

12  yards  +  9  feet. 
81 


82  ALGEBRA  —  FIRST  COURSE  [viii,§44 

That  is,  we  add  the  parts  of  the  distances  measured  by  one  kind  of 
units  together,  then  add  the  parts  measured  by  another  kind  of  units 
together,  and  combine  the  results  to  get  the  measurement  of  the  sum 
of  the  lines. 

If  we  regard  the  inch  as  the  unit  of  measure  in  this  exer- 
cise, and  the  yard  and  the  foot  as  multiples  of  this  unit, 
our  exercise  becomes: 

Add  a  line  3  •  36  inches  +  7  •  12  inches  long  to  a  line  9  •  36 
inches  +  2  •  12  inches  long.  Adding  as  before  the  sum  is 
12  .  36  inches  +  9-12  inches.     So  that 

(9  .  36  +  2  .  12)  +  (3  •  36  +  7  •  12)  =  12  .  36  +  9  .  12. 

Expressing  this  more  generally  by  letting  y  stand  for  the 
number  of  inches  in  a  yard  and  /  stand  for  the  number  of 
inches  in  a  foot,  we  have 

(92/  +  2/)  +  (3  2/  +  7/)=12  2/  +  9/. 

Example  2.  Suppose  that  a  Une  measures  9  yards  and  1  foot  and 
4  inches.  Let  y  stand  for  a  length  of  one  yard,  /  for  a  length  of  one 
foot,  and  i  for  a  length  of  one  inch,  each  measured  in  terms  of  another 
unit,  as  a  centimeter.     Then  the  length  of  the  line  in  centimeters  is 

expressed  by 

92/+/  +  4i. 

This  is  a  trinomial. 

Suppose  that  a  second  Une  measures  3  yards  and  2  feet  and  6  inches. 
Its  length  is  then  expressed  by 

Now  the  sum  of  the  lengths  of  these  lines  will  be  expressed  by 

(9y+/  +  4i)  +  (32/  +  2/  +  6i), 

and  we  know  that  we  can  get  this  sum  by  adding  yards  to  yards,  feet 
to  feet,  and  inches  to  inches.     So  we  have 

(92/  +/  +  4i)  +  (3y  +  2/  +:6z)  =  12 2/  +  3/  +  10  I. 

In  this  way  any  two  polynomials  are  added  together,  by  adding  like 
terms. 

The  difference  of  the  lengths  of  our  two  lines  is  expressed  by 

(92/+/  +  4i)-^3y  +  2/  +  6i). 


VIII.  §  441 


ADDITION  OF  POLYNOMIALS 


83 


and  again  we  know  that  this  difference  can  be  found  by  subtracting 
terms  measured  in  like  units.    So  we  have 

Let  the  student  check  this  by  adding. 

In  the  same  way  we  might  have  such  expressions  as  the  following; 
state  each  of  these  as  a  problem  in  measurement,  and  check  each  result. 

(9y  +  4i)  +  (62/  +  2/-8i)  =  151/ +  2/-  4  i. 

{9y  +  4i)  -(Qy  +  2f  +  Qi)  =    Sy-2f-2i. 

iSy-3f  +  5i)  -ilQy  +  2f-Si)  =-2y-5f  +  Si. 

Make  up  for  yourself  several  more  exercises  of  this  kind. 
Geometric  Illustration. 

Add:  (2m-3/) +  (-3m  +  4/-2) +  (-m  +  6/). 
Solviion: 

I 


ill! 

IS 
1 

5 

1     1     1 

0 

1     !     1 

1 

1  1  1  1 

10 

1     1     1     1 

-3  m 

1       )       1       r 

-5  m 

[ 

1  ^ 

~? 

1  )a 

1   "ijii 

3  m 

^^^ 

I     1     1 

\ 

1  ;  1 

j        ' 

1      1      1      1 

-A/ 

\r'^ 

0 

3f 

hf 

i" 

i     5/ 
! 

^ 

^ 

K— 

Algebraic  expression: 

(2w-3/)  +  (-3m  +  4/-2)  +  (-m  +  6/)  =-2m  +  7/  -  2. 

In  the  diagram  the  measuring  line  for  /  might  have  been  laid  off  on 
the  original  scale  line,  but  it  is  much  less  confusing  to  draw  a  new  scale 
to  count  by.  The  student  is  again  cautioned  not  to  forget  that  all  of 
the  segments  representing  the  numbers  are  supposed  to  lie  on  the 
original  scale  line.  In  order  not  to  forget  that  the  thing  we  are  in- 
terested in  is  the  distance  we  are  from  the  point  where  we  began  to 
count,  it  is  well  to  mark  that  distance  on  the  scale  line,  every  time  we 
change  the  measuring  line,  as  shown  in  the  figure. 

Check :  Let  m  =  3,  /  =  1^  or  f ,  as  shown  in  the  figure. 


84  ALGEBRA  —  FIRST  COURSE  [VIII.  §44 

Substituting  in  the  given  expression, 

(2.3-3-f)  +  (-3.3  +  4.|-2)  +  (-3  +  6.f) 
=  (6  -  4i)  +  (-  9  +  6  -  2)  +  (-  3  +  9) 
=  H  +  C-5)+6 

Substituting  in  the  answer, 

-2.3  +  7-li-2=-6  +  10^-2 


Since  we  get  2^  from  each  expression,  the  expressions  are  equal. 
This  is  also  the  result  in  the  figure. 

In  the  following  exercises  make  diagrams  for  the  addition 
of  some  of  them;  add  the  others  and  check  all. 

It  is  of  great  importance  that  fractional  and  negative 
numbers  be  used  in  at  least  part  of  the  checks,  since  it  im- 
presses the  fact  that  the  letters  given  may  have  either 
integral  or  fractional  values  and  that  these  may  be  either 
positive  or  negative;  —  a  may  be  either  a  positive  or  a 
negative  number  according  to  the  value  of  a. 

Exercises. 

1.  Add  2r-sto5r-2s. 

2.  Add  Qx  +  5y-7to2x-7y  +  5. 

3.  Add  3a- 166 +  2  to  4a +  96  + 4c. 
Add: 

4.  (2r-3p+-6)  +  (4r  +  8p-2)  +  (6r  +  9). 

6.    (- 6a  +  5c -46 -6)  + (-8a +  66)  + (16a +  9). 

6.  (3r-6s  +  10  +  20  +  (6r  +  7^-24)  +  (-5s-12r). 

7.  (7s  +  50  +  (-4a  +  60+(7^-4a  +  95)+(-19i-4s). 

In  the  subtraction  of  polynomials  the  same  idea  enters 
as  in  addition.  We  subtract  the  terms  separately  and  add 
the  results  of  the  subtractions. 

Example.    Subtract  2  r  —  3  s  from  r  —  4  s  +  2  <. 
Writing  in  algebraic  form: 

(r-4s  +  20  -  (2r-3s)  =? 


VIII.  §441  SUBTRACTION  OF  POLYNOMIALS  85 

The  first  question  is  —  What  must  be  added  to  2  r  to  get  r  ?    The 
answer  is  —  r.     The  second  question  is  —  What  must  be  added  to  —  3  s  to 
get  —  4s?    The  answer  is  —  s.     The  third  question  is  —  What  must 
be  added  to  0  to  get  2  t  ?    The  answer  is  2  t.    So  we  have 
(r-4s  +  2  0-(2r-3s)=-r-s  +  2«i 

Check :  The  check  is  to  add  the  result  to  the  second  number  to  see 
if  you  get  the  first. 

(2r-3s) +  (-r -s  +  20  =r-4r  +  2«. 

It  would  be  well  to  make  a  drawing  of  a  few  of  the  checks.  The 
student  must  be  careful  to  do  his  checking  conscientiously;  that  is,  he 
must  actually  do  the  adding  or  he  will  be  led  to  make  many  blunders. 

8.  Subtract  7  r  +  2  s  from  2  r  +  6  s. 

9.  Subtract  -Zs-\-2t  from  7 s  -  bt. 

10.  Subtract  —2a  —  Zh  from  a  +  6. 
In  the  following  do  the  work  indicated. 

11.  (2  c  +  7  d  -  6)  -  (5  c  -  2  e  +  7). 

12.  (-5^-36- 10) -(-2^-46-  19). 

13.  (76-4a-6c  +  4)  -  (-96 -4a  +  6c- 4). 

14.  (2  a  -  3  c)  +  (5  a  +  16  c)  -  (3  a  -  8  c). 

15.  (6-6(^-4a  +  9)  +  (7r  +  86- 17d) 

-  (4r-7(^  +  3a-  10). 

16.  (-7/+8sf-10/i-p)  +  (5/-8/i  +  /i-10p) 

-(-5/-9sr-6). 

17.  (7r-6^-9s- 15)  +  (-4r-12i+12s) 

-  (9r  +  2s-  17). 

18.  Write  the  supplements  of  the  following  angles: 

a  degrees;  2  a  —  6  degrees;  a  +  3  6  degrees;  2  a  —  4  6  +3 
degrees;  2  r  radians;  J  tt  —  r  radians;   If  x  +  2  r  radians. 

19.  Write  the  complements  of  the  angles  of  Exercise  18. 

20.  Write  the  complement  of  angle  a,  then  write  the 
supplement  of  this  complement.  Draw  this.  Do  you  get 
the  same  answer  in  your  drawing  as  in  your  algebra? 

21.  Write  the  complement  of  the  angle  a,  then  write  the 
supplement  of  this  complement,  then  write  the  complement 
of  this  supplement.  Draw  and  see  if  you  get  the  same 
result. 


86  ALGEBRA  —  FIRST  COURSE  [Vlll.§46 

45.  Rules  for  Adding  and  Subtracting  Polynomials. 

(1)  To  add  two  polynomials,  add  like  terms,  each  term  to  be 
taken  with  the  sign  before  it. 

(2)  To  subtract  two  polynomials,  subtract  like  terms,  each 
term  to  be  taken  with  the  sign  before  it. 

Also,  since  subtracting  a  quantity  is  equivalent  to  adding 
its  negative,  we  may  use  the  following  rule  for  subtracting 
polynomials. 

(3)  To  subtract  one  polynomial  from  another,  change  the 
signs  of  all  the  terms  of  the  polynomial  to  be  subtracted  and  add 
the  result  to  the  other  polynomial. 

Try  Rule  3  in  some  of  the  preceding  exercises. 

46.  Removal  of  Parentheses.  From  what  you  have  now 
learned  about  polynomials  you  will  easily  see  that  such  equa- 
tions as  the  following  are  true. 

-\-{a  —  b  —  c)=a  —  b  —  c. 
—  {a  —  b  —  c)  ——  a-\rb  -{-  c. 

Rule.  When  a  polynomial  is  inclosed  in  parentheses  pre- 
ceded by  a  positive  sign,  the  parentheses  may  be  omitted. 

When  a  polynomial  is  inclosed  in  parentheses  preceded  by 
a  negative  sign,  we  may  omit  the  parentheses  only  if  we  at  the 
same  time  change  the  sign  of  each  term  of  the  polynomial. 

We  shall  apply  this  to  an  example  involving  several  signs 
of  aggregation  (§8). 

Example.    Remove  all  signs  of  aggregation  from  the  expression 
-  {a -[(6  + 3d)  -(4e-/)]  +  (26-7d  +  3e)j. 

Here  we  may  first  remove  the  parentheses  inside  the  brackets  [  ]; 
we  then  have 

-  5a-[6  +  3d-4e+/]  +  (26-7d  +  3e)i. 

Now  remove  brackets  after  changing  the  sign  of  each  term  inclosed; 
also  omit  the  parentheses;  we  then  have 

-Sa-6-3d  +  4e-/  +  2&-7d  +  3ei. 

Combining  similar  terms,  we  have 

-  |a  +  &-10d  +  7e-/}. 


VIII.  5  47]  SUMMARY  .  87 

Now  remove  braces  after  changing  the  signs  of  all  terms  inclosed; 
we  finally  have 

-a-h  +  10d-7e+f. 

In  this  work  it  should  be  noticed  that  brackets  and  braces 
are  merely  different  forms  of  parentheses. 

Exercises.  Simplify  the  following,  removing  all  signs  of 
aggregation. 

1.  [a+{c-d)-  (3a  +  2d)]. 

2.  [-  (2  m  -  n)  +  (5  m  +  6  n)  -  (8  m  -  3  n)  +  5  m]. 

3.  -[2x+(4:y-7x)-  (Sy  +  Sx)-  (5x-2y)]. 

4.  -[-iu  +  2v-Sw)-\-{3u-2v  +  w)-u  +  Sv-^w], 
6.  -\[-(7x-2y)-{-{Sx-4.y)]-[2x-5y-(3x-y)]\. 

N.B.  Whenever  the  expression  inclosed  in  parentheses 
can  be  simplified  by  a  reduction  or  combination  of  its  terms, 
this  should  be  done  first. 

47.  Summary. 

The  amount  of  a  quantity  may  be  measured  by  comparing 
it  with  different  units.  The  units  may  be  of  the  standard 
kind  as  used  in  arithmetic,  or  they  may  be  general  units. 

When  only  one  length  is  used  this  measurement  is  ex- 
pressed by  means  of  one  term  and  the  expression  is  called 
a  monomial. 

When  two  different  lengths  are  used  in  the  same  measure- 
ment, the  result  is  expressed  by  two  terms  and  the  expression 
is  called  a  binomial. 

When  three  different  lengths  are  used  in  the  same  meas- 
urement, the  result  is  expressed  by  means  of  three  terms 
and  the  expression  is  called  a  trinomial. 

In  general,  when  two  or  more  lengths  are  used  in  the  meas- 
urement of  a  quantity,  there  are  two  or  more  terms  in  the 
expression  of  the  measurement  and  the  expression  is  called 
a  polynomial. 

To  add  polynomials.  Write  the  polynomials  in  parentheses 
with  the  plus  sign  between  them.    Add  the  like  terms  and 


88  ALGEBRA  —  FIRST  COURSE  [Vlll,§47 

express  the  answer  with  different  terms,  one  following  the 
other,  connected  by  the  sign  obtained  by  the  addition. 

To  subtract  polynomials.  Write  the  polynomials  in  paren- 
theses, the  one  to  be  subtracted  following  the  one  from  which 
it  is  to  be  subtracted,  with  the  minus  sign  between  them. 
Subtract  the  like  terms  in  the  manner  that  you  learned  to 
subtract  monomials,  and  write  the  result  with  the  different 
terms,  one  following  the  other,  connecting  them  by  the 
sign  obtained  from  the  subtraction. 

Removal  of  parentheses.  If  a  parenthesis  is  preceded  by  a 
plus  sign,  the  parenthesis  may  be  removed  without  changing 
the  signs  of  the  terms  in  the  parenthesis.  If  a  parenthesis 
is  preceded  by  a  minus  sign  the  parenthesis  may  be  removed, 
provided  the  sign  of  each  term  in  the  parenthesis  is  changed. 

Problems  in  Addition  and  Subtraction. 

1.  On  a  winter  day  the  temperature  at  noon  was  twice 
the  temperature  at  6  a.m.  and  the  temperature  at  6  p.m.  was 
two-thirds  of  the  temperature  at  noon.  The  sum  of  the 
three  was  65  degrees.     Find  the  temperature  at  each  time. 

2.  The  income  from  a  certain  business  during  the  second 
year  was  double  the  income  during  the  first  year;  during 
the  third  year  it  again  doubled.  The  total  income  was 
$4500.     What  was.  the  income  each  year  ? 

3.  The  volume  of  a  cylindrical  cup  in  cubic  inches  is  3 
times  the  volume  of  a  conical  cup  of  the  same  height  and 
base.  Using  these  as  measures  it  is  found  that  2  measures 
of  the  first  and  5  of  the  second  just  fill  a  gallon  can.  What 
is  the  volume  of  each  cup  ? 

4.  If  2  a  dollars  and  5  h  dimes  are  taken  from  4  a  dollars 
and  3  b  dimes  the  remainder  is  $4.60.  Also  3  b  dollars 
equal  15  6  dimes.  Find  the  numbers  for  which  a  and  b 
stand. 

5.  Show  by  drawing  that 

(2  +  3)  -  (8  -  5  -  6)  =  (2  +  3)  +  (-  8  +  5  +  6). 


VIII.  1 47]  SUMMARY  89 

6.  Three  children  are  digging  dandelions.  One  digs  at 
the  rate  of  a  dozen  per  hour;  the  second  at  the  rate  of  6 
dozen  per  hour;  the  third  at  the  rate  of  c  dozen  per  hour. 
The  first  morning  the  first  works  4  hours,  the  second  3  hours, 
and  the  third  5  hours.  The  second  morning  the  first  works 
5  hours,  the  second  3  hours,  and  the  third  5  hours.  How 
many  more  dandelions  were  dug  the  second  day  than  the 
first? 

7.  There  are  four  sizes  of  shot.  The  largest  size  weighs 
s  grams,  the  second  weighs  r  grams,  the  third  p  grams.  On 
one  pan  of  the  scales  are  7  of  the  first  kind,  15  of  the 
second,  18  of  the  third,  and  25  of  the  fourth.  On  the  other 
pan  are  13  of  the  first,  6  of  the  second,  30  of  the  third,  and 
4  of  the  fourth.  How  many  grams  weight  are  on  the  first 
pan  ?  How  many  are  on  the  second  ?  How  much  must  be 
added  to  the  weight  on  the  second  pan  to  have  the  scales 
balance? 


CHAPTER  IX 
MULTIPLICATION   AND   DIVISION    OF    POLYNOMIALS 
48.   Multiplication  of  a  Polynomial  by  a  Monomial. 

Example  1.  The  length  of  a  rectangle  is  5  inches,  its  width  is  3 
inches;  the  length  of  a  second  rectangle  is  5  inches  and  its  width  is 
4  inches.     What  is  the  sum  of  their  areas? 


^111        II  5 


Area  of  the  first  rectangle  is  5  •  3  square  inches. 
Area  of  the  second  rectangle  is  5  •  4  square  inches. 
Then  the  sum  of  these  areas  is  (5  •  3  +  5  •  4)  =35  square  inches. 
This  is  equivalent  to  the  area  of  a  rectangle  whose  dimensions  are 
6  by  7,  as  shown  in  the  figures  above. 
That  is,  we  have 

5.  3  +  5.  4  =  5  (3 +  4) 
=  5-7. 

In  exactly  the  same  way,  if  the  dimensions  of  the  first  rectangle 
are  a  by  6  and  the  dimensions  of  the  second  rectangle  are  a  by  c,  the 
sum  of  their  areas  is 

a'h  -\-a'  c, 

but  this  must  be  equivalent  to  a  single  rectangle  whose  dimensions 
are  a  by  (,6  +  c\     That  is,  we  always  have 

a6  +  oc  =  a  (6  +  c). 

Example  2.  If  '3  boys  pull  a  sled  forward  with  a  force  of  30  pounds 
and  3  other  boys  pull  the  sled  forward  with  a  force  of  35  pounds  each, 
what  is  the  total  pull  on  the  sled? 

90 


IX.  §48]  MULTIPLICATION  OF  POLYNOMIALS  91 

Solution:  We  can  get  the  result  in  two  different  ways. 
First:  Total  pull  of  the  first  3  boys  is  3  •  30  pounds. 

Total  pull  of  the  other  3  boys  is  3  •  35  pounds. 
Therefore  the  whole  is  the  sum  of  these, 

(3  .  30  +  3  •  35)  pounds  =  (90  +  105)  pounds 
=  195  pounds. 

Second:  Taking  one  boy  from  each  set  of  three,  their  combined  pull 
is  (30  +  35)  pounds.  Since  there  are  three  such  pairs  of  boys,  their 
total  pull  is 

3  (30  +  35)  pounds  =  3-65  pounds 
=  195  pounds. 
Hence  we  have 

3  (30  +  35)  =  3  .  30  +  3  .  35.     Why? 
This  is  another  illustration  of  the  rule 

a  (6  +  c)  =  ah  -\-  ac. 

State  this  rule  In  words. 

It  is  easy  to  illustrate  this  rule  when  some  of  the  numbers  are  nega- 
tive. 

Example  3.  If  3  boys  pull  a  sled  forward  with  a  pull  of  30  pounds 
each  and  if  3  boys  pull  the  sled  backward  with  a  force  of  35  pounds 
each,  what  is  the  effective  pull  on  the  sled? 

Solution: 

Use  +  for  pull  forward  and  —  for  pull  backward. 

First:  3  •  30  +  3  •  (-  35)  =  90  +  (-  105) 

=  -  15. 

That  is,  the  net  effect  is  a  backward  pull  of  15  pounds. 
Second:   Take  the  boys  in  pairs,  one  pulling  forward  and  the  other 
pulling  backward.     The  net  effect  of  the  pulling  of  each  pair  is 
30 +  (-35)  =-5. 
Since  there  are  three  such  pairs,  the  total  effect  is 

3  [30  +  (-35)]  =  3-  (-5) 
=  -  15. 

Since  the  expressions  in  the  first  and  second  solutions  stand  for  the 
same  thing,  namely  total  pull,  we  have 

3  [30  +  (-  35)]  =  3  .  30  +  3  (-  35). 

Using  letters,  suppose  n  boys  pull  the  sled  forward  with  a  force  of 


92  ALGEBRA  —  FIRST  COURSE  UX.  §  48 

p  pounds  each,  and  n  other  boys  pull  it  backward  with  a  force  of  q 
pounds  each,  then  in  the  first  solution  we  would  have 

n-v  +  n(-  g); 
In  the  second  solution  we  would  have 

n[v\-{-  q)l 

Here  again  we  must  have 

n[p  +  i-q)]  =n'p+n{-q), 

which  is  the  same  form,  except  for  a  difference  in  the  letters,  as 

a  (6  +  c)  =  a6  +  ac. 

If,  in  the  last  example,  both  sets  of  boys  pull  backward,  show  that 
the  total  pull  may  be  expressed  in  either  of  the  forms 

n  K-p)  +  (-g)]  or  n{-p)  +n(-g), 

so  that  we  have 

n[{-p)  -f-  i-q)]  =n{-p)  +n(-g). 

Another  example. of  this  sort  is  the  following. 

Example.  A  man  who  can  row  at  the  rate  of  Vr  miles  an  hour  in 
still  water  enters  a  stream  which  has  a  velocity  of  Vs  miles  an  hour. 
How  far  can  he  row  against  the  stream  in  one  hour?  How  far  can  he 
row  in  three  hours? 

Solution:  In  this  example  if  we  regard  the  direction  in  which  the  man 
is  going  as  positive,  then  the  direction  in  which  the  stream  is  flowing  is 
negative. 

Therefore  the  distance  gone  in  an  hour  is 

Vr  +  i-Va)  =Vr-  V,. 

The  distance  gone  in  3  hours  is 

3  {Vr  —  Va). 

But  the  man  would  be  just  as  far  at  the  end  of  3  hours  if  it  were 
possible  for  him  to  stop  the  stream  and  he  rowed  for  3  hours  in  still 
water,  then  he  stopped  rowing  and  allowed  the  stream  to  flow  at  its 
accustomed  rate  and  pull  him  along  for  3  hours.  In  the  first  case  the 
two  velocities  are  acting  simultaneously,  while  in  the  latter  case  they 
would  be  acting  consecutively. 

The  algebraic  expression  of  the  first  case  is  as  given  above.  For 
the  latter  case  it  would  be 

3t;r-3t;,. 


IX,  §49]  MULTIPLICATION  OF  POLYNOMIALS  93 

Therefore, 

3  {Vr  —  Va)  =  SVr  —  S  Vg. 

This  is  illustrated  graphically  below. 


Vr 

I — . — '■    '     1     1 — .->^ 


;  S(Vt-Vs) ^ 


-l-*l 


J I t_ 


-V5 

1 \ L 


3V, 


SV^SV, 


>|  -SVa 


49.  Distributive  Law.  The  equation  a{b  -\-c)  =  ah  -}- ac 
and  the  similar  equations  just  considered  express  what  is 
known  as  the  distributive  law.  This  law  asserts  that  the 
product  of  a  single  number  by  the  sum  of  two  numbers  is 
identical  with  the  sum  of  the  products  of  the  first  number 
by  the  other  two  numbers  taken  singly. 

As  to  which  of  the  two  ways  this  shall  be  written  will 
depend  entirely  upon  the  use  we  are  going  to  make  of  it  in 
our  work. 

Exercises. 

1.  A  girl  earned  I  cents  an  hour  helping  in  the  laboratory, 
and  c  cents  an  hour  for  correcting  papers.  She  spent  4 
hours  a  day  in  the  laboratory,  and  3  hours  a  day  marking 
papers.  How  much  did  she  earn  during  a  month  of  20 
days?  First  write  the  algebraic  expression  for  the  entire 
amount  earned  in  one  day,  then  write  it  for  the  amount 
earned  in  20  days.  Also  write  the  amount  earned  for  lab- 
oratory work  during  the  month,  then  the  amount  earned 
for  correcting  papers  during  the  month,  and  add  the  two 
amounts  together,  as  the  total  amount  earned.  These  two 
expressions  for  the  amount  earned  during  the  month  are 
equal  to  each  other.     Write  this  equality. 

2.  A  teacher  prepared  for  her  class  b  bottles  of  a  solution, 
each  of  which  contained  w  grams  of  water  and  15  grams  of 


94  ALGEBRA  — FIRST  COURSE  rix,§50 

salt.  How  many  grams  of  the  solution  were  there?  Write 
the  expression  for  this  in  two  different  ways  placing  the 
equality  sign  between  them. 

3.  If  a  student  ^ends  h  dollars  a  month  for  board  and  r 
dollars  a  month  for  room  rent  for  6  months,  how  much  does 
he  spend?  Write  the  equality  of  the  two  different  algebraic 
expressions  for  the  answer  to  this. 

4.  From  each  of  a  bottles  of  water,  each  containing  g 
grams,  an  average  of  3  centigrams  evaporated.  How  many 
centigrams  were  left  in  the  bottles? 

6.  If  a  boy  earns  w  dollars  a  day  and  pays  c  cents  a  day 
for  his  board,  how  much  has  he  by  the  end  of  the  month? 

6.  A  measuring  stick  is  3  c  feet  and  2  b  inches  long. 
When  applied  to  a  line  it  goes  3|  times.  How  long  is  the 
line? 

7.  A  measuring  line  lacks  I  feet  of  being  20  feet  long. 
When  applied  to  a  line  it  goes  /  times,  how  long  is  the 
line? 

8.  A  cubic  centimeter  of  gold  weighs  w  grams  and  a 
cubic  centimeter  of  lead  weighs  I  grams  less.  What  is  the 
weight  of  I  cubic  centimeters  of  lead? 

Write  problems  leading  to  the  following  expressions; 
change  the  letters  to  any  suitable  to  the  problem,  and 
write  the  two  algebraic  expressions  of  the  answers  equal  to 
each  other. 

9.   3  (a +  4).  12.   5(8a  +  7&). 

10.   5  (-  2  a  +  3).  13.   a  (a-  1). 

U.   7  (6  a-  1).  14.   a(a  +  b). 

15.   b(a  +  2h  -  c). 

50.  The  Commutative  Law  of  Addition.  This  law  as- 
serts that  the  value  of  the  sum  of  two  numbers  does  not 
depend  upon  the  order  in  which  the  numbers  are  taken 
when  they  are  added. 

Algebraically  expressed,  a  +  b  =b-\- a,  no  matter  what  the 
sign  of  a  and  6. 


IX,  §51]  MULTIPLICATION  OF  POLYNOMIALS  95 

Geometrically  expressed, 


J 1 I I 1 I I         '  t 


Thus  we  see  that  if  the  two  numbers  are  positive  the  sums 
are  the  same. 

In  like  manner :  a -{■  (—  b)  =—  h  -\-  a. 

0 

t I I I I \ I I I I 


-b 

. 

a. 

^1 

^i 

^-" 

-6 

1 

-      a+(-6)  [ 

1 

1 

a 

In  like  manner:  —  a -\-h  =  h  -\-  (—  a). 

0 


^ — r- 

' s ^^ 


■  a+b 

k 


In  like  manner :  ^  a  -{-  (—  h)  =  —  h  -\-  (—  a). 


-a+(>&)  U 1 — 


-b-i-i-a)  K" 


-6 


51.  The  Commutative  Law  of  Multiplication.     This  law 
asserts  that  the  value  of  the  product  of  two  numbers  does 


96  ALGEBRA  —  FIRST  COURSE  [ix.§52 

not  depend  upon  the  order  in  which  the  numbers  are  taken 
when  they  are  multiphed. 

Algebraically  expressed:  db  =  ba,  no  matter  what  the  sign 
of  a  and  b. 

Sylvester's  illustration  of  the  commutative  law  for  multi- 
pHcation.  Take  a  baskets  containing  b  apples  each.  Now 
since  there  are  b  apples  in  one  basket,  in  a  baskets  there 
are  a  times  b  apples  or  ab  apples.  Take  an  apple  from  each 
of  the  baskets  and  place  in  a  new  basket;  that  basket  will 
have  a  apples  in  it.  Again  take  an  apple  from  each  of  the 
original  baskets,  and  you  will  have  another  new  basket 
with  a  apples  in  it.  If  you  keep  on  this  way,  you  will  at 
last  transfer  all  the  apples  in  the  original  baskets  to  b  new 
baskets,  and  there  will  be  a  apples  in  each  basket.  Since 
there  are  a  apples  in  each  basket,  in  b  baskets  there  will  be 
b  times  a  apples  or  ba  apples. 

Therefore,  ab  =  ba. 

52.  Exponents.  The  area  of  a  rectangle  is  the  product 
of  its  two  dimensions.  If  these  dimensions  are  a  and  b, 
the  area  is  ab.  If  the  rectangle  is  a  square,  whose  dimen- 
sion is  a,  then  its  area  is  a  •  a.  The  volume  of  a  cube  whose 
edge  is  a,  is  a  •  a  •  a. 

To  shorten  the  writing  of  such  products,  we  use  the  fol- 
lowing symbols. 

In  place  of  a  •  a  write  a^. 

In  place  of  a  •  a  •  a  write  a^. 

In  place  of  a  •  a  •  a  •  a  write  a^. 

In  place  of  a  •  a  •  a  •  a  •  a  write  a^,  and  so  on. 

These  are  read  a  square,  a  cube,  a  fourth  power,  a  fifth 
power,  and  so  on. 

The  little  number  written  above  and  to  the  right  of  an- 
other number  shows  how  many  times  the  other  number  is 
to  be  used  as  a  factor;  the  little  number  is  called  the  ex- 
ponent, and  the  other  number  is  called  the  base. 


IX,  §52]  MULTIPLICATION  OF  POLYNOMIALS  97 

We  also  write 

a%  =  a*  a  '  b;    a%^  =  a*  a-  a  'h  'b;    and  so  on. 

Exercises.  As  drill  to  become  familiar  with  this  notation, 
write  the  following;  make  use  of  the  exponent.  Read  what 
you  have  written. 

1.  3  aaabb.  5.  4|  hhhhk  •  4|  hhkkst. 

2.  7  •  7  mrrmn.  6.  4  •  4  •  4  •  a  •  4  bbba. 

3.  8  r  •  8  sstttr.  7.  2  r  •  2  r  •  5  r. 

4.  i'iccdddfd.  8.  -  7  a  -  (- 7  a)  ab. 

Write  the  following  without  making  use  of  the  exponent. 

9.   9  •  3^  m  •  3  r^m^n.  11.   32  r^s%^  •  (-  S^rh%^). 

10.   ^cd'UcdJ.  12.    (-3)2(-3a)3. 

13.  Write  the  answers  to  Exercises  9,  10,  11,  12,  making 
use  of  the  exponent. 

14.  Write  as  products  of  prime  factors,  20,  72,  36,  75, 
98,  312. 

15.  Calculate  23 .  3^ .  52. 

Meaning  of  word  coefficient.  In  expressions  of  the  kind 
with  which  we  have  just  been  dealing,  we  call  the  figures 
the  numerical  factors  and  the  letters  the  literal  factors. 
The  numerical  factors  are  usually  written  first,  that  is,  pre- 
ceding the  literal  factors. 

When  we  wish  to  refer  to  one  particular  factor,  we  call 
the  other  factors  coefficients  of  this  one.  Thus,  in  the  ex- 
pression 4  abCf  if  we  are  interested  in  c,  then  4  ab  is  its  co- 
efficient; 4  be  is  the  coefficient  of  a;  4:ae  is  the  coefficient 
of  6;  4  6  is  the  coefficient  of  ae;  4  is  the  coefficient  of  abc; 
abe  is  the  coefficient  of  4.  Coefficient  means  co-factor.  How- 
ever, the  numerical  factor  is  referred  to  unless  otherwise 
stated.  If  you  were  called  upon  to  give  the  coefficient  in 
the  term  3  abcj  you  would  say  that  it  is  3. 


98  ALGEBRA  —  FIRST  COURSE  [IX,  5  53 

63.   Multiplication  of  One  Polynomial  by  Another. 

Example  1.  A  teacher  in  a  laboratory  prepared  for  a  class  w  bottles, 
each  containing  w  grams  of  water  and  15  grams  of  salt.  The  class 
used  13  bottles.     How  many  grams  of  the  solution  were  there  left? 

Solution:  We  can  compute  this  in  two  different  ways,  and  give 
algebraic  expressions  for  the  different  ways  of  computing.  The  easier 
way  to  compute  it  is  to  subtract  the  number  of  bottles  used  from  the 
number  of  bottles  prepared,  and  then  compute  the  number  of  grams  of 
solution  in  those  left.     The  algebraic  expression  for  this  is 

{w  —  13)  {w  +  15)  =  the  number  of  grams  of  the  solution  left; 

or  we  can  compute  the  number  of  grams  of  solution  that  we  had  at 
first  and  from  this  subtract  the  number  used,  and  thus  arrive  at  the 
number  of  grams  left.     The  algebraic  expression  for  this  is 

w  {w  -\-  15)  —  13  (ly  +  15)  =  the  number  of  grams  left; 

therefore       (w  -  13)  (w  +  15)  =  w  {w  +  15)  -  13  (w;  +  15);  (Why?) 
or,  writing  in  the  other  form,      =  {w^  +  15  w)  —  {ISw  +  195).' 
Subtracting  as  indicated,  =  w^  -\-  2  w  —  195,  the  number  of  grams 

left. 

Check  by  letting  w  =  60.  Substituting  in  the  original  problem: 
If  a  teacher  in  a  laboratory  prepared  for  a  class  60  bottles  each 
containing  60  grams  of  water  and  15  grams  of  salt,  there  were  in  each 
bottle  the  sum  of  60  grams  and  15  grams  which  is  75  grams.  There 
were  at  first  60  bottles,  but  the  class  used  13  bottles;  there  were  then 
47  bottles  left.  Since  there  were  75  grams  in  each  bottle,  in  47  bottles 
there  would  be  47  times  75  grams  which  is  3525  grams.  This  is  the 
number  of  grams  left  after  the  class  used  13  bottles. 

Substituting  w  =  QO  in  the  answer,  which  is  supposed  to  be  the 
number  of  grams  left,  we  have 

wj2  +  2  wj  -  195  =  602  +  2  •  60  -  195 
=  3600  +  120  -  195 
=  3525. 

Since  this  is  the  number  which  we  got  by  our  arithmetic  solution, 
w^  +  2w  —  195  must  be  the  correct  number  of  grams  left  after  the 
class  used  13  bottles. 

Example  2.  A  man  who  can  row  at  the  rate  of  Vr  miles  an  hour  in 
still  water,  rows  up  a  stream  which  flows  at  the  rate  of  Vg  miles  an  hour. 
He  rows  m  hours  in  the  morning  and  a  hours  in  the  afternoon.  How 
many  miles  does  he  row  upstream  during  the  day? 


IX,  §53]  MULTIPLICATION  OF  POLYNOMIALS  99 

Solution:  As  before,  we  can  compute  this  in  two  different  ways  to 
arrive  at  the  correct  result.  First  add  together  the  number  of  hours 
he  rows  in  the  morning  and  in  the  afternoon,  finding  the  number  of  hours 
he  rows  during  the  day;  then  multiply  the  number  of  miles  he  progresses 
per  hour  by  this.  Second,  find  the  number  of  miles  he  rows  in  the 
morning  and  the  number  of  miles  he  rows  in  the  afternoon ;  add  these, 
obtaining  the  number  of  miles  he  rows  during  the  day.  The  algebraic 
expressions  of  these  two  computations  are  equal  to  each  other.  This 
gives  the  equation 

(w  +  a)  (Vr  -Vs)  =m  {Vr  -Vs)  +a  {Vr  -  Vs); 
multiplying  on  the  right,  =  {mvr  —  mva)  +  {avr  —  avg) ; 

adding  as  indicated,  =  mVr  —  mva  +  aVr  —  avg, 

the  number  of  miles  the  man  rows  up  the  stream  during  the  day. 

Check  by  letting  m  =  S,  a  =  2,  Vr  =  5,  Vg  =2.  Substituting  in  the 
original  problem: 

If  a  man  who  can  row  at  the  rate  of  5  miles  an  hour  in  still  water, 
rows  up  a  st'-eam  which  flows  at  the  rate  of  2  miles  an  hour,  he  will 
go  up  the  stream  at  the  rate  of  3  miles  an  hour.  Since  he  rows  3  hours 
in  the  morning  and  2  hours  in  the  afternoon,  he  rows  during  the  day 
the  sum  of  3  hours  and  2  hours  which  is  5  hours.  Since  he  advances  3 
miles  in  1  hour  and  rows  for  5  hours,  he  will  advance  5  times  3  miles, 
which  is  15  miles. 

Substituting  in  the  answer: 

mVr  —  mVa  -\-  avr  —  ava  =  3«5  —  3«2  +  2«5  —  2»2 
=  15-6  +  10-4 
=  15. 

Since  this  is  the  number  we  got  from  our  arithmetic  solution 
mAr  —  mva  +  avr  —  avs  must  be  the  correct  number  of  miles  he  rows. 

Exercises.     Solve  the  following  as  illustrated  above. 

1.  In  an  alloy  of  metal  there  are  in  each  cubic  centimeter 
g  grams  of  one  metal  and  4  grams  of  another.  In  one 
brick  of  the  metal  there  are  g  cubic  centimeters,  and  in 
another  there  are  15  cubic  centimeters.  How  many  grams 
do  the  two  bricks  together  weigh? 

2.  Two  boys  each  agree  to  work  for  w  dollars  a  day,  out 
of  which  25  cents  is  taken  for  dinner.  If  one  works  t  days 
and  the  other  5  days,  how  much  do  they  together  receive? 


100  ALGEBRA  —  FIRST  COURSE  [ix,§53 

3.  How  much  more  does  the  boy  who  works  t  days  receive 
than  the  boy  who  works  5  days? 

4.  Two  men  start  out  to  walk.  One  walks  t  days  a 
week  for  2  weeks  and  the  other  walks  m  days.  If  they  both 
walk  d  miles  in  the  morning  and  t  miles  in  the  afternoon, 
how  much  farther  does  the  one  go  than  the  other? 

5.  There  are  two  pieces  of  land.  One  is  square  and  the 
other  is  rectangular.  The  rectangular  piece  has  a  length 
6  meters  longer  than  the  side  of  the  square,  and  a  width  11 
meters  shorter  than  the  side  of  the  square.  How  many 
more  square  meters  are  there  in  the  area  of  the  square  than 
in  the  area  of  the  rectangle?     • 

6.  The  length  of  one  side  of  a  square  is  s  meters.  A  rec- 
tangle is  1  meter  more  in  length  and  1  meter  less  in  width, 
what  is  the  difference  in  the  area  of  the  two  figures?  Which 
is  the  larger? 

7.  The  side  of  a  square  is  s  meters.  A  rectangle  is  3 
meters  more  in  length  and  3  meters  less  in  width.  How 
much  more  is  the  area  of  the  square  than  that  of  the 
rectangle? 

8.  The  side  of  a  square  is  s  meters.  A  rectangle  is  5 
meters  more  in  length  and  5  meters  less  in  width.  How 
much  more  is  the  area  of  the  square  than  the  area  of  the 
rectangle? 

9.  The  side  of  a  square  is  s  meters.  A  rectangle  is  7 
meters  more  in  length  and  7  meters  less  in  width.  How 
much  greater  is  the  area  of  the  square  than  the  area  of  the 
rectangle? 

10.  The  side  of  a  square  is  s  meters.  A  rectangle  is  12 
meters  more  in  length  and  12  meters  less  in  width.  How 
much  greater  is  the  area  of  the  square  than  that  of  the 
rectangle? 

11.  In  Exercises  8,  9,  10, 11  and  12,  are  the  perimeters  the 
same  in  each  case?  How  do  the  areas  of  the  different  rec- 
tangles compare?    Draw  them,  arranging  them  in  the  order 


IX.  §54]  DIVISION  OF  POLYNOMULS  101 

of  their  size  (use  a  millimeter  instead  of  meter  as  the  unit). 
If  rectangles  have  equal  perimeters,  which  is  the  greatest? 

Write  problems  leading  to  five  of  the  following.  Solve 
all,  expressing  the  results  in  the  different  algebraic  forms  as 
in  the  preceding  exercises: 

.12.  (m  +  2)  (m  +  5).  16.  i2k-7)&k-  15). 

13.  (^1  -  «  (si  +  S2).  17.  (5r-/i)(3r  +  2/i). 

14.  (r  +  6)  (r  -  2).  18.  (c  -  1)  (c^  +  c  +  1). 
16.  (p-8)(p-7).  19.  (62  +  6  a6)  (2  62  -  3a6). 

20  (m  +  3  rs)  {m}  —  3  mrs  +  9  rh^). 

21.  (a +  3  6)  (a +  3  6). 

22.  (m^n  +  2  a%)  {m^n  +  2  a%). 

23.  (-4-6a26)(-4  +  6a26). 

24.  (-6 a;^^/  +  2  xy^)  (-Qx^y-{-2 xy^). 

25.  i-d'-2cd^){d^-2cd^). 

26.  (a:2  +  5)  (x2  +  X  -  30). 

54.  Division  of  Polynomials. 

Example  1.  A  man,  who  can  row  at  the  rate  of  5  miles  an  hour  in 
still  water,  rows  upstream  against  a  2-mile  current.  He  rows  24  miles 
the  first  day  and  15  miles  the  second.     How  many  hours  did  he  row? 

Solution :  He  advances  in  one  hour  a  distance  of  (5  —  2)  miles. 
He  rowed  (24  +  15)  miles. 

Therefore  the  time  he  rowed  is 

f±^  =  f      (Wh.V) 
=  13  hours. 

Here  we  can  simplify  our  first  expression  by  combining  the  numbers, 
and  afterwards  by  dividing  out. 

Suppose  the  same  problems  to  be  stated  in  a  literal  form. 

A  man,  who  can  row  at  the  rate  of  r  miles  an  hom*  in  still  water, 
rows  upstream  against  the  current  flowing  c  miles  an  hour.  He  rows 
n  miles  the  first  day  and  m  miles  the  second.     How  long  did  he  row? 

Solution:  He  advances  in  one  hour  a  distance  (r  —  c)  miles.  He 
rows  (n  +  m)  miles.     Therefore  the  time  used  in  rowing  is 

^-t^  hours.    (Why?) 


102  ALGEBRA  —  FIRST  COURSE  [ix.§54 

We  cannot  now  simplify  the  result  further  so  we  leave  it  as  it 
stands.  It  is  the  indicated  quotient  of  one  binomial,  namely  (n  +  m),  by 
another  binomial,  namely  (r  —  c).  Such  indicated  quotients  are  called 
fractions. 

Example  2.  The  cost  of  seeding  a  piece  of  ground  was  4  dollars 
per  acre,  the  cost  of  cultivating  was  2  dollars  per  acre,  and  the  cost  of 
harvesting  was  3  dollars  per  acre.  The  total  cost  was  $270.  How 
many  acres  were  there? 

Solution:  The  total  cost  per  acre  is  (4  +  2  +  3)  dollars. 

Therefore  the  number  of  acres  is 

270  270  ^3„_ 


4+2+3         9 

If  the  cost  of  seeding  is  s  dollars,  the  cost  of  cultivating  is  c  dollars, 
the  cost  of  harvesting  is  h  dollars,  and  if  the  total  cost  is  a  dollars,  we 
shall  have  the  number  of  acres 


s  +  c  +  /i 

We  shall  study  such  form  farther  on  in  the  chapter  on 
factoring. 

Problems  in  Division  of  Polynomials. 

1.  If  a  man  can  row  at  the  rate  of  r  miles  an  hour  in  still 
water,  and  if  he  rows  downstream  in  a  current  flowing  at 
the  rate  of  c  miles  an  hour,  how  many  hours  will  it  take  him 
to  go  d  miles?  If  he  rows  upstream,  how  long  does  it  take 
him  to  go  d  miles  ? 

2.  If  it  costs  a  cents  to  set  the  type  for  a  printed  circular, 
and  if  it  costs  h  cents  a  hundred  for  the  printing,  and  c  cents 
a  hundred  for  the  paper,  how  many  circulars  can  be  made 
for  n  dollars  ? 

3.  To  construct  an  office  building  costs  for  each  story  / 
dollars  for  the  floor,  w  dollars  for  the  walls,  p  dollars  for  the 
partitions.  The  total  cost  of  the  building  is  c  dollars,  which 
includes  h  dollars  for  the  basement  and  r  dollars  for  the  roof. 
How  many  stories  are  there  ? 

4.  A  room  is  a  feet  long  and  h  feet  wide,  and  another  room 
is  c  feet  long  and  d  feet  wide.     It  costs  k  dollars  to  lay  a 


IX.§S5]  SUMMARY  103 

floor  in  these  two  rooms.     What  is  the  price  of  the  floor  per 
square  foot  ? 

6.  If  a  man  loans  p  dollars  to  one  person  and  q  dollars  to 
each  of  two  others,  and  receives  i  dollars  interest  for  the 
two  years,  what  is  the  rate  of  interest? 

6.  If  the  temperature  at  a  certain  place  was  d,  e,  and  / 
respectively  for  three  consecutive  days,  what  was  the  aver- 
age temperature  for  the  time? 

7.  If  three  rectangles  having  the  same  altitude  and  bases 
which  are  6,  c,  and  d  respectively,  are  added  so  that  their 
bases  are  in  a  straight  line,  their  area  is  a  square  inches. 
What  is  their  altitude  ? 

8.  A  cube,  a  cylinder,  a  cone,  and  a  sphere  of  iron  weigh 
Vi  Q)  '"'i  s  grams  respectively,  and  have  a  combined  volume  of 
V  cubic  centimeters.     What  is  the  density  of  iron? 

9.  The  momentum  of  a  boy  riding  his  bicycle  is  m  mile- 
pounds  per  hour.  The  weight  of  the  boy  was  Wi  pounds  and 
the  weight  of  the  bicycle  was  W2  pounds.  What  was  the 
rate  per  hour  at  which  they  were  moving?  (Momentum  in 
mile-pounds  per  hour  =  weight  in  pounds  X  speed  in  miles 
per  hour.) 

55.   Summary. 

Commutative  law  of  addition.  This  law  asserts  that  the 
value  of  the  sum  of  two  numbers  does  not  depend  on  the 
order  of  summation. 

Commutative  law  of  multiplication.  This  law  asserts  that 
the  value  of  the  product  of  two  numbers  does  not  depend  on 
the  order  of  multiplication. 

A  positive  integral  exponent  of  a  number  is  a  small  number 
written  to  the  right  and  a  little  above  the  number  to  show 
how  many  times  the  number  is  used  as  a  factor.  The  num- 
ber to  which  the  exponent  is  attached  is  called  the  base. 

A  coefficient  of  a  number  is  a  cofactor  with  that  number. 

To  multiply  a  polynomial  hy  a  monomial.    Write  the  num- 


104  ALGEBRA  —  FIRST  COURSE  [ix.§55 

bers  as  factors  —  the  monomial  followed  by  the  polynomial 
inclosed  in  parentheses.  Make  this  equal  to  the  product 
of  each  term  of  the  poljrnomial  multiplied  by  the  monomial, 
the  partial  products  being  connected  by  the  proper  sign  to 
form  the  complete  product. 

To  multiply  a  polynomial  hy  a  polynomial.  Write  the  two 
polynomials  in  parentheses  as  factors.  Multiply  the  mul- 
tiplicand by  each  term  of  the  multiplier,  writing  the  partial 
products  in  parentheses  connected  by  the  proper  sign.  Add 
or  subtract  as  indicated  by  the  sign. 


CHAPTER  X 
PROBLEMS   LEADING   TO    SIMPLE   EQUATIONS 

56.  Problems. 

Example  1.  Apparatus  needed  —  a  pair  of  balances  and  some  shot 
(any  substance  of  which  you  can  have  a  number  of  the  same  size  and 
weight  will  do).  Place  a  quantity  of  shot  in  one  pan  of  the  balances, 
and  a  different  quantity  in  the  other  together  with  weights  enough  to 
balance.  Find  the  weight  of  each  shot.  See  that  at  all  times  the 
scales  are  balanced. 

Now  the  process  which  you  go  through  in  the  actual  use  of  the  scales 
in  order  to  find  the  weight  of  each  shot  can  be  expressed  in  algebraic 
symbols. 

Suppose  you  had  placed  23  shot  in  one  pan,  and  12  in  the  other,  and 
then  found  it  necessary  to  place  22  gram  weights  on  the  second  pan  in 
order  to  balance  the  scales.  Since  you  are  interested  in  the  weight  of 
one  shot  only,  the  thing  that  you  would  do  in  the  actual  operation  of 
weighing  would  be  to  take  out  of  both  pans  the  same  number  of  shot, 
until  you  had  shot  only  in  one  pan.  Then  you  would  count  the  num- 
ber of  shot  left  in  the  pan  and  the  number  of  gram  weights  left  in  the 
other  pan,  and  from  this  compute  the  weight  of  one  shot.  We  shall 
now  represent  this  in  algebraic  language. 

Let  w  =  the  number  of  grams  in  the  weight  of  each  shot; 

then  23  w  =  the  number  of  grams  in  the  weight  of  23  shot, 

and  12  w  =  the  number  of  grams  in  the  weight  of  12  shot. 

Therefore      23  w  =  the  number  of  grams  in  one  pan, 
while    12  w  +22  =  the  number  of  grams  in  the  other  pan. 

Since  these  two  amounts  balance  each  other,  they  are  equal  to  each 
other.    Therefore  we  write 

2Sw  =  12w  +  22. 

Now,  just  as  we  did  in  the  actual  weighing,  we  can  take  out  12  w 
from  each  pan  and  have 

llw?  =  22. 
105 


106  ALGEBRA  —  FIRST  COURSE  ix,§56 

Since  w  is  multiplied  by  11  to  be  equal  to  22,  and  2  is  multiplied  by 
11  to  be  equal  to  22,  therefore, 

w  =2. 

Therefore,  since  w  stood  for  the  weight  of  each  shot,  each  shot  must 
weigh  2  grams. 

Check:  We  can  test  the  correctness  of  this  result  by  computing  the 
weight  in  each  pan. 

If  1  shot  weighs  2  grams,  23  shot  will  weigh  23  •  2  or  46  grams, 
which  is  the  weight  in  one  pan. 

12  shot  will  weigh  12  •  2  or  24  grams.  To  this  add  the  22  grams, 
and  we  have  46  grams  as  the  weight  on  the  other  pan. 

The  two  amounts  are  equal  and  hence  balance  each  other,  therefore 
our  answer  must  be  correct. 

1.  Place  different  quantities  of  shot  in  each  pan  and  then 
balance  the  scales  by  placing  gram  and  fractional  gram 
weights  on  each  pan.  Proceed  to  find  the  weight  of  each 
shot.  See  that  at  each  moment  the  scales  are  kept 
balanced. 

State  this  experiment  in  words;  represent  in  algebraic 
symbols;  solve  and  check. 

2.  A  boy  found  that  if  he  placed  2  iron  balls  of  an  equal 
size  on  one  pan  of  the  scales,  he  must  place  50  gram  weights 
on  the  other,  in  order  to  balance  them.  What  is  the  weight 
of  each  ball? 

3.  In  weighing  a  stone  block  in  the  laboratory,  a  student 
found  that  if  he  placed  a  20-gram  weight  on  the  side  of  the 
balances  with  the  stone,  it  would  balance  a  100-gram  weight 
on  the  other  side.  What  is  the  weight  of  the  stone?  Ex- 
plain this  without  using  symbols,  then  solve  by  means  of 
symbols. 

4.  If  5  marbles  of  equal  size  and  5  gram  weights  balance 
2  marbles  and  40  gram  weights,  what  is  the  weight  of  each 
marble? 

5.  If  your  book  and  20  grams  balance  625  grams,  what 
is  the  weight  of  your  book? 


X,§56]  SIMPLE  EQUATIONS  107 

Example  2.  If  21  cc.  of  marble  weigh  56.7  grams,  what  is  the 
density  of  marble?     That  is,  what  is  the  weight  of  one  cc.  of  marble? 

Solution: 

Let  d  =  the  density  of  marble. 

Then  21  d  =  the  number  of  grams  in  the  weight  of  21  cc. 

But  56.7  =  the  number  of  grams  in  the  weight  of  21  cc, 

therefore       21d  =  5Q.7, 
and  d  =  2.7,  the  density  of  marble. 

Check :  If  one  cubic  centimeter  of  marble  weighs  2.7  grams,  21  cc. 
will  weigh  21  times  2.7,  which  is  56.7.  This  is  the  amount  that  the 
example  states  that  it  should  weigh.     Therefore  the  result  is  correct. 

6.  How  many  cubic  centimeters  are  there  in  a  brass  ball 
which  weighs  109.2  grams?     The  density  of  brass  is  8.4. 

7.  If  you  place  on  one  pan  of  the  scales  a  piece  of  wood 
and  another  f  as  large,  they  together  balance  16  lbs.  in 
weights  placed  on  the  other  pan.  What  is  the  weight  of 
each  piece  of  wood? 

8.  The  density  of  platinum  is  2.17  more  than  the  density 
of  gold.  Two  bars,  one  containing  20  cc.  of  platinum  and 
the  other  containing  20  cc.  of  gold,  together  weigh  826  grams. 
What  is  the  density  of  platinum  ?  of  gold  ? 

9.  A  cubic  centimeter  of  ebony  weighs  1  gram  more  than 
a  cubic  centimeter  of  cork.  A  piece  of  cork  containing 
16  cc.  and  a  piece  of  ebony  containing  7  cc.  placed  together 
on  one  pan  of  the  scales  balance  11.14  gram  weights  placed 
on  the  other  pan.  What  is  the  density  of  cork?  of  ebony? 
What  is  the  weight  of  each  piece  ? 

10.  Iron  is  4  times  as  heavy  as  ivory.  14  ivory  balls  and 
6  iron  balls  of  the  same  dimensions  are  placed  on  the  scales 
and  are  found  to  balance  456  gram  weights.  What  is  the 
weight  of  each  ball?  What  is  the  density  of  ivory  if  each 
ball  contains  6f  cc?    What  is  the  density  of  iron? 

11.  If  20  shot  and  a  5-gram  weight  together  balance  4 
shot  and  29  grams,  what  is  the  weight  of  each  shot  ? 

12.  If  2  iron  blocks  of  the  same  size,  together  with  4  oz. 
weights,  just  balance  a  block  f  as  large  as  one  of  them 


108  ALGEBRA  —  FIRST  COURSE  [x.  §  56 

together  with  12  oz.  weights,  what  is  the  weight  of  each 
block? 

13.  A  boy  placed  in  one  pan  of  the  scales  110  shot  and  in 
the  other  50  shot  and  25  grams  in  weights.  He  then  re- 
moved from  the  first  pan  enough  to  make  the  two  pans 
balance.  He  found  afterward  by  calculation  that  he  had 
removed  shot  enough  to  weigh  15  grams.  What  was  the 
weight  of  each  shot? 

14.  The  weight  of  one  cubjc  centimeter  of  copper  is  1.1 
grams  more  than  that  of  the  same  amount  of  iron.  5  iron 
balls  weigh  278.3  grams  more  than  3  copper  balls  of  the 
same  size.  What  is  the  density  of  iron,  if  each  ball  contains 
23  cc. ?    What  is  the  density  of  copper? 

15.  The  density  of  lead  lacks  4.3  of  being  2  times  that  of 
iron.  6  iron  balls  weigh  1.6  grams  more  than  4  lead  balls  of 
the  same  size.  What  is  the  weight  of  each  iron  ball  if  each 
ball  contains  1  cc.  ?     What  is  the  weight  of  each  lead  ball  ? 

16.  An  iron  pail  covered  with  tin  weighs  1.1366  kilograms. 
There  are  125  cc.  of  iron  and  22  cc.  of  tin.  If  iron  has  a 
density  of  .5  more  than  tin,  what  is  the  density  of  iron  and 
of  tin? 

17.  If  3  cc.  of  a  substance  weigh  9  grams,  what  does 
1  cc.  weigh.  If  16  cc.  of  a  substance  weigh  38  grams,  what 
is  the  density  of  the  substance?  If  84  cc.  of  a  substance 
weigh  21  grams,  what  is  the  density  of  the  substance?  If 
6  cc.  weigh  1  gram,  what  is  the  density  of  the  substance  ?  If 
V  cc.  weigh  m  grams,  what  is  the  density  of  the  substance? 

The  answer  to  this  last  question  is  usually  expressed  in 

the  form 

,      m 
d  =  —• 

V 

This  can  be  used  as  a  formula  by  which  to  solve  other 
exercises.     Using  this  as  a  formula  solve  the  following: 

18.  A  block  of  lead  in  the  shape  of  a  cube  whose  edge  is 
10  cm,  weighs  11,300  grams.    What  is  the  density  of  lead? 


X.§57]  SIMPLE  EQUATIONS  109 

19.  A  block  of  marble  6  dm.  wide,  6  dm.  thick  and 
1.5  m.  long  weighs  1,468,800  grams.  What  is  the  density  of 
marble  ? 

20.  A  block  of  silver  2  cm.  long,  c  cm.  wide  and  c  cm.  thick 
weighs  21  grams.     What  is  the  density  of  silver? 

21.  The  mass  of  a  substance  is  7  grams  and  its  volmne  is 
10  cc.     What  is  its  density? 

22.  The  mass  of  a  substance  is  7ab  grams  and  its  volume 
is  14  b  cc.     What  is  its  density? 

23.  Solve  —  not  using  formula: 

Alcohol  is  .8  as  heavy  as  water.  A  bottle  containing 
354.3  cc.  of  alcohol  and  118.1  cc.  of  water  weighs  401.54 
grams.  If  the  weight  of  the  glass  is  not  included  in  this 
weight,  what  is  the  density  of  alcohol  ?  of  water  ? 

24.  From  the  Table  of  Densities  on  p.  35,  write  three 
problems  leading  to  equations  and  solve. 

57.  Formation  of  Equations.  In  all  of  these  exercises 
you  will  have  noticed  that  certain  amounts  are  known  and 
certain  amounts  are  unknown.  You  will  also  have  noticed 
that  we  get  the  values  of  those  that  are  unknown  through 
their  relations  to  those  that  are  known.  Those  quantities 
that  are  given  by  experiment  and  measurement  are  called 
known  quantities.  Those  whose  values  are  to  be  found 
through  their  relation  to  the  known  quantities  are  called 
unknown  quantities. 

Whenever  we  have  enough  data  on  the  relations  of  one 
unknown  quantity  to  known  quantities  to  be  able  to  form 
two  sets  of  symbols  standing  for  the  same  amount,  we  can 
place  the  sets  equal  to  one  another,  and  solve  for  the 
unknown  quantity  as  we  have  done  in  the  preceding 
exercises. 

The  expression  formed  by  placing  these  two  sets  of  syra- 
bols  equal  to  one  another  is  called  an  equation. 

The  truth  which  we  employ  in  forming  the  equation  is 
called  an  axiom.     It  is  usually  worded: 


110  ALGEBRA  —  FIRST  COURSE  [X.  §  58 

Things  which  are  equal  to  the  same  thing  are  equal  to 
each  other. 

An  axiom  is  a  statement  whose  truth  we  assume  without 
proof. 

58.  Solution  of  Equations.  Having  formed  our  equa- 
tion, we  further  assume  that  we  may  add  equal  amounts^to 
both  sides  of  the  equation  without  destroying  the  equaUty, 
or  that  we  may  subtract  equal  amounts  from  both  sides  of 
the  equation  without  destroying  the  equality.  Thus  we  get 
an  equation  with  the  unknown  terms  on  one  side  of  the 
equation  and  the  known  terms  on  the  other.  Likewise  we 
assume  that  both  sides  of  an  equation  may  be  multiplied  or 
divided  by  equal  quantities  without  destroying  the  equality. 
This  leads  to  the  solution  of  the  equation. 

So  we  have  the  following  five  axioms  by  means  of  which 
we  solve  our  equation: 

I.  Things  equal  to  the  same  thing  are  equal  to  each 

other. 
II.  If  equals  be  added  to  equals,  the  results  are  equal. 
ni.  If  equals  be  subtracted  from  equals,  the  results  are 

equal. 
IV.  If  equals  be  multiplied  by  equals,  the  results  are 

equal. 
V.  If  equals  be  divided  by  equals,  the  results  are  equal. 

Example  1.  A  boy  riding  his  wheel  starts  for  his  home  20  miles 
away.  He  rode  2  miles  farther  the  second  hour  than  he  did  the  first, 
and  3  miles  farther  the  third  hour  than  he  did  the  first,  and  reached  his 
destination.     How  far  did  he  ride  each  hour? 

Solution:  Let  d  =  the  number  of  miles  the  boy  rode  the  first  hour. 
Then  d  -\-  2  =  the  number  of  miles  he  rode  the  second  hour, 

and  d  +  3  =  the  number  of  miles  he  rode  the  third  hour. 

Then  3  cZ  +  5  =  the  number  of  miles  he  rode; 

but  20  =  the  number  of  miles  he  rode. 

Therefore  3  d  +  5  =  20.     (Things  which  are  equal  to  the  same  thing  are 
equal  to  each  other.) 


X.§58]  SIMPLE  EQUATIONS  111 

Now  subtract  5  from  each  side  of  the  equation;  we  have 

Sd  =  15.     (If  equals  are  subtracted  from  equals,  the  results 
are  equal.) 
Then       d  =  5,  (If  equals  are  divided  by  equals,  the  results  are  equal.) 
and  d  +  2  =  7,  (If  equals  are  added  to  equals,  the  results  are  equal.) 
and  d  +  3  =  8.  (For  the  same  reason.) 

^o  the  boy  rode  5  miles  the  first  hour,  7  miles  the  second  hour  and  8 
miles  the  third  hour. 

Check:  If  a  boy  rode  his  wheel  at  the  rate  of  5  miles  an  hour  the 
first  hour  and  at  the  rate  of  7  miles  an  hour  the  second  hour,  he  rode 
2  miles  farther  the  second  hour  than  the  first.  If  he  rode  8  miles  the 
third  hour,  he  rode  3  miles  farther  the  third  hour  than  the  first.  If 
he  rode  5  miles  the  first  hour,  7  miles  the  second  hour  and  8  miles  the 
third  hour,  he  rode  altogether  (5  +  7  +  8)  miles,  which  equals  20  miles, 
the  entire  distance.  Therefore  since  these  answers  check  each  state- 
ment of  the  original  example,  they  must  be  correct. 

Problems. 

1.  Two  boys  returning  from  a  fishing  trip  wished  to 
divide  the  fish  caught  equally  between  them.  Upon  count- 
ing them  the  one  who  had  caught  the  most  said  to  the  other, 
*'  If  I  had  6  more,  I  should  have  2  times  as  many  as  you  have. 
I  will  give  you  11,  then  we  shall  have  an  equal  number." 
How  many  fish  did  each  have  ? 

2.  A  class  committee  appointed  to  look  up  the  price  of  a 
piece  of  statuary  reported  that  the  money  in  the  treasury 
was  but  50  cents  more  than  half  as  much  as  was  needed  for 
the  purchase,  and  that  it  would  be  necessary  for  the  class  to 
raise  62  dollars  more  in  order  to  buy  the  piece  selected. 
How  much  did  the  piece  of  statuary  cost  ? 

3.  Two  boys  pulled  a  small  wagon  in  opposite  directions, 
the  forward  pull  being  three  times  the  backward  pull.  The 
result  was  a  forward  pull  of  74  lbs.  What  was  the  amount 
and  direction  of  each  boy's  pull  ? 

4.  Three  men  on  a  building  are  trying  to  raise  a  beam 
with  ropes.  One  of  the  men  exerts  a  force  of  f  as  many 
pounds  as  the  strongest  of  the  three,  and  the  other  a  force  of 
^  as  many  pounds  as  the  strongest.    They  succeed  in  lifting 


112  ALGEBRA  —  FIRST  COURSE  [x.§58 

the  beam  which  weighs  470  pounds.     What  is  the  number  of 
pounds  force  of  each  man's  pull  ? 

5.  A  young  man  wishing  to  earn  money  to  pay  his  board 
while  attending  school  decided  to  build  fires  in  furnaces  at 
private  residences  in  the  neighborhood.  He  found  that  the 
number  of  engagements  that  he  could  make  for  this  kind  of 
work  was  just  }  of  the  number  of  cents  that  he  asked  per 
day  at  each  residence,  and  that  if  he  would  lower  his  price 
5  cents  per  day  at  each  residence  he  could  get  2  more  en- 
gagements, and  thus  earn  15  cents  more  per  day.  How 
much  did  he  ask  per  residence? 

6.  A  lady  arranging  a  flower  bed  found  that  she  had 
plants  enough  to  arrange  exactly  in  the  form  of  a  square, 
but  if  she  attempted  to  arrange  them  in  the  form  of  a  bed 
with  4  more  in  the  length  and  2  less  in  the  width,  she  would 
lack  16  of  having  enough  plants.  How  many  plants  has 
she? 

7.  Of  three  lines  one  is  6  cm.  longer  than  f  of  the  length 
of  another,  and  the  third  is  7f  cm.  less  than  the  sum  of  the 
other  two,  while  their  combined  length  is  9  cm.  What  is 
the  length  of  each  line? 

8.  There  are  two  angles  whose  sum  is  97  degrees.  If  13 
degrees  be  subtracted  from  the  larger  and  added  to  the 
smaller,  the  two  angles  will  be  equal.  What  is  the  size  of 
each  angle? 

Example  2.  A  solution  of  alcohol  and  water  is  95%  strong.  How 
much  water  must  be  added  to  have  it  75%  strong? 

Remark.  By  95%  strong  we  mean  that  .95  of  any  selected  amount 
or  part  is  alcohol,  and  that  5%  is  water. 

Then  in  starting  an  exercise  of  this  kind,  unless  some  definite  amount 
is  stated,  we  can  speak  of  one  part,  a  part  being  any  convenient  amount. 

SoltUion: 

Let  w  =  the  number  of  parts  of  water  that  must  be  added  to 

one  part  of  the  given  solution. 
Then  1  -{■  w  =  the  number  of  parts  in  the  new  solution. 

.75  (1  -\-  w)  =  the  number  of  parts  of  alcohol. 


X,§58]  SIMPLE  EQUATIONS  113 

But  .95  =  the  number  of  parts  of  alcohol,  smce  the  amount  of 

alcohol  is  not  changed. 
.-.      .75  (1  +  w)  =  .95.     Why? 

w  =  ^5,  the  number  of  parts  of  water  that  must  be  added 
to  reduce  a  95%  solution  to  a  75%  solution. 
Check  as  in  previous  problems. 

9.  Upon  examining  a  bottle  of  rose-water  and  glycerine 
which  she  had  just  bought,  a  lady  said  to  the  druggist, 
"There  seems  to  be  too  much  glycerine  in  this."  He  an- 
swered "I  thought  you  said  that  you  wished  equal  parts." 
The  lady  answered  "No,  I  said  that  I  wished  35%  glycerine." 
How  much  rose-water  must  be  added  if  the  lady  is  willing 
that  it  be  added  to  the  four  ounces  that  are  in  the 
bottle? 

10.  A  solution  of  sulphuric  acid  and  water  is  33  J  %  sul- 
phuric acid.  How  much  water  must  be  added  to  have  it  10% 
sulphuric  acid? 

11.  In  a  solution  of  iodine  and  alcohol,  there  is  6.25%  of 
iodine.  How  much  alcohol  must  be  added  in  order  to  have 
1%  iodine? 

12.  A  2-ounce  bottle  of  peroxide  is  95%  strong.  This  is 
too  strong  for  general  use.  How  much  water  must  be  added 
to  make  a  10%  solution? 

13.  A  pint  of  ammonia  is  purchased  for  cleaning  purposes. 
It  is  a  90%  solution.  This  is  too  strong.  How  much  water 
must  be  used  if  a  50%  solution  is  desired  ? 

14.  If  sea-water  is  12%  salt,  how  much  water  must  be 
evaporated  in  order  to  have  it  90%  salt? 

16.  Write  three  problems  relating  to  solutions  with  which 
you  are  acquainted.  Metals  in  alloy  may  be  used.  Solve 
your  problems. 

16.  If  one  angle  of  a  triangle  is  15  degrees  more  than 
another,  and  a  third  is  2  times  the  sum  of  these  two,  what  is 
the  number  of  degrees  in  each  angle  of  the  triangle  ? 

17.  If  one  of  the  angles  of  a  triangle  is  15  degrees  more 
than  3  times  another,  and  J  of  the  difference  between  these 


114  ALGEBRA  —  FIRST  COURSE  ix.§58 

two  is  equal  to  J  of  the  third  angle,  what  is  the  number 
of  degrees  in  each  angle  of  the  triangle  ? 

18.  If  you  subtract  5  degrees  from  the  number  of  degrees 
in  the  first  angle  of  a  triangle  and  multiply  the  remainder 
by  3,  you  will  get  the  number  of  degrees  in  the  second  angle 
of  the  triangle.  The  third  angle  is  4  times  the  second. 
What  is  the  number  of  degrees  in  each  angle  of  the  tri- 
angle ? 

19.  The  second  side  of  a  triangle  is  5  times  the  first,  and 
the  third  is  5  cm.  longer  than  the  first.  If  the  third  is  sub- 
tracted from  the  second,  the  difference  will  lack  2  cm.  of 
being  equal  to  the  first.     What  is  the  length  of  each  side  ? 

20.  One  side  of  a  triangular  lot  is  200  feet  longer  than  an- 
other, and  the  third  side  is  2  times  this  one.  It  requires 
1100  feet  of  fence  to  enclose  the  lot.  What  is  the  length  of 
each  side? 

21.  A  triangle  has  two  of  its  sides  equal.  If  1  were  added 
to  the  number  of  centimeters  in  the  length  of  one  of  them, 
the  sum  would  be  3  times  the  number  of  centimeters  in  the 
length  of  the  third  side.  If  3  cm.  were  subtracted  from  each 
of  the  equal  sides  and  added  to  the  third  side,  the  triangle 
would  be  equilateral.  What  is  the  length  of  each  side  of 
the  triangle  ? 

22.  If  one  angle  of  a  triangle  is  ^  tt  radians  more  than  f 
of  another,  and  the  third  is  equal  to  4  times  the  difference 
between  these  two,  what  is  the  number  of  radians  in  each 
angle  of  the  triangle  ? 

Write  problems  concerning  a  triangle  leading  to  the  fol- 
lowing equations: 

23.  s-f(s-M)  +  (2s-2)=27. 

24.  5a  -  30  =  180.  25.   2ia  +  Jtt  =  t. 

Example  3.     If  an  angle  is  2f  times  its  complement,  what  is  the 
number  of  degrees  in  the  angle?  in  the  complement? 
Solviion: 
Let  o  =  the  number  of  degrees  in  the  angle. 


X.§58]  SIMPLE  EQUATIONS  115 

Then        90  —  o  =  the  number  of  degrees  in  the  complement.     Why? 
2f  (90  —  a)  =  the  number  of  degrees  in  the  angle.     Why? 
2f  (90  -a)  =a.     Why? 
.*.   240  —  21  a  —  a,  doing  the  multiplication  indicated. 

240  =  3|  a,  adding  2|  a  to  both  members.     Why? 

a  =  -52-.     Why? 

=  Vr,  multiplying  numerator  and  denominator  by  3, 
=  65xr,  number  of  degrees  in  the  angle. 
90  —  a  =  24/r,  number  of  degrees  in  the  complement  of  the 
angle. 

Check:  Since  the  angles  are  complementary  their  sum  should  equal 

90  degrees.    (65r\  +  24r\)  degrees  is  90  degrees.     The  angle  is  as 

/^f'"5_  7  on 

many  times  its  complement  as  ^^t^j  which  is  ^^=7;,  which  is  2f .    Since 

24j f  z70 

the  answers  obtained  satisfy  all  the  conditions  of  the  problem,  they 

must  be  correct. 

26.  If  an  angle  is  200  degrees  more  than  2  times  its  com- 
plement, what  is  the  size  of  the  angle  and  of  its  complement  ? 

27.  If  —  15  degrees  be  added  to  |  of  the  supplement  of 
an  angle,  the  sum  will  be  equal  to  the  angle.  What  is  the 
number  of  degrees  in  the  angle?  in  the  supplement  of  the 
angle  ? 

28.  How  many  radians  are  there  in  an  angle  if  its  comple- 
ment is  3  times  its  supplement  ? 

29.  Two  angles  are  complementary.  If  x\  tt  radians  be 
added  to  one  and  subtracted  from  the  other,  the  two  angles 
will  be  equal.     What  is  the  number  of  degrees  in  the  angles  ? 

30.  Write  three  problems  concerning  complementary  an- 
gles and  supplementary  angles.  Solve  to  see  that  they  are 
correct. 

31.  One  acute  angle  of  a  right-angled  triangle  is  IJ  times 
the  other.  What  is  the  number  of  radians  in  each  of  the 
angles  ? 

32.  Of  two  adjacent  sides  of  a  parallelogram  one  is  |  as 
long  as  the  other.  The  perimeter  is  49  cm.  What  is  the 
length  of  each  side  ? 


116  ALGEBRA  —  FIRST  COURSE  IX,§58 

33.  Of  the  six  angles  of  a  hexagon,  A,  B,  C,  D,  E,  F, 
angle  5  is  5  degrees  more  than  2  times  angle  A ;  angle  C  is 
14  degrees  more  than  ^  of  angle  B;  angle  D  is  2  times  as 
large  as  angle  C;  angle  E  lacks  10  degrees  of  equalling  J 
angle  A ;  angle  F  equals  the  difference  between  angle  A  and 
angle  E.  The  sum  of  the  angles  of  a  hexagon  is  8  right 
angles.  Find  the  number  of  degrees  in  each  angle  of  the 
hexagon. 

34.  The  sum  of  two  angles  is  150  degrees;  f  of  the  smaller 
equals  J  of  the  larger.  What  is  the  number  of  degrees  in 
each  angle? 

35.  In  a  triangle  whose  sides  are  a,  h,  c,  if  the  side  c  con- 
tained 1  unit  more  it  would  be  just  J  of  side  a;  if  it  were  1 
unit  less  it  would  be  J  the  length  of  side  h.  Furthermore, 
if  one  unit  b^  subtracted  from  side  a  and  added  to  side  h, 
sides  a  and  h  would  be  equal.  Find  the  length  of  each  side 
of  the  triangle. 

36.  Three  men,  A,  B,  C,  on  a  building,  are  trying  to  raise 
a  beam  with  ropes.  A  pulls  with  a  force  of  f  as  many 
pounds  as  C,  and  B  pulls  with  a  force  of  f  as  many  pounds 
as  C.  The  beam  weighs  471  pounds.  With  what  force 
does  each  man  pull  if  we  regard  a  downward  pull  as  positive  ? 

37.  The  density  of  wrought  silver  is  7.945  more  than  that 
of  agate.  A  silver  ornament  set  with  agate  contains  8  cc. 
of  each,  and  weighs  110.6  gm.  Find  the  density  of  silver 
and  of  agate. 

38.  A  glass  bottle  containing  alcohol  is  closed  with  a 
cork.  There  are  23  cc.  of  glass  in  the  bottle  and  50  cc.  of 
alcohol;  the  cork  measures  5  cc.  The  whole  weighs  100.12 
grams.  If  a  cubic  centimeter  of  alcohol  and  one  of  cork 
together  weigh  1.08  grams,  and  the  density  of  glass  is  .7 
less  than  4  times  that  of  alcohol,  what  is  the  density  of  each 
substance  ? 

39.  Two  men  start  from  a  point  200  miles  apart  and  travel 
toward  each  other,  one  at  the  rate  of  5  miles  an  hour,  and 


X.§58]  SIMPLE  EQUATIONS  117 

the  other  at  the  rate  of  10  miles  an  hour.    After  how  many 
hours  will  they  meet  ? 

40.  ^,  traveling  at  the  rate  of  20  miles  a  day,  has  4  days 
start  of  B,  traveling  at  the  rate  of  26  miles  a  day,  in  the  same 
direction.     After  how  many  days  will  B  overtake  A  ? 

41.  The  circumferences  of  the  front  wheel  and  of  the  hind 
wheel  of  a  wagon  are  2  and  3  yards  respectively.  What 
distance  has  the  wagon  moved  when  the  front  wheel  has 
made  10  revolutions  more  than  the  hind  wheel? 

Hint.  —  Let  r  =  number  of  revolutions  the  front  wheel 
will  make. 

42.  The  sum  of  two  digits  of  a  number  is  12.  If  the 
digits  be  interchanged,  the  resulting  number  will  be  equal 
to  the  original.     What  is  the  number  ? 

43.  The  sum  of  the  two  digits  of  a  number  is  12.  If 
the  digits  be  interchanged,  the  resulting  number  exceeds  the 
original  number  by  |  of  the  original  number.  What  is  the 
number  ? 

44.  The  sum  of  two  digits  of  a  number  is  11.  If  the 
digits  are  interchanged,  the  resulting  number  is  45  less  than 
the  number.     What  is  the  number  ? 

45.  Write  three  problems  on  numbers  and  their  digits 
and  solve. 

46.  One  boy  says  to  another:  "Think  of  a  number;  add 
7;  double  result;  take  away  8;  tell  me  your  answer  and  I 
will  tell  you  the  number  thought  of.''     How  can  he  do  it? 

47.  Make  up  a  problem  like  number  46. 

48.  In  sending  packages  by  parcel  post  for  the  5th  zone, 
the  amount  of  increase  on  each  pound  is  2  cents  less  than  the 
cost  of  the  first  pound.  The  cost  of  11  pounds  is  7  cents 
more  than  8  times  the  cost  of  the  first  pound.  What  is  the 
cost  of  each  pound  in  this  zone? 

49.  In  the  3rd  zone  the  cost  for  the  first  pound  is  2  cents 
less  than  2  times  the  additional  price  per  pound.  The  price 
for  11  pounds  is  4  cents  more  than  7  times  the  cost  for  the 


118  ALGEBRA  —  FIRST  COURSE  [x,  §  58 

first  pound.     What   is  the   cost  for  each  pound   in   this 
zone? 

Note.  —  In  solving  a  problem  it  is  not  necessary  that  the  letter  stand 
for  a  number  of  single  units  of  the  thing  called  for;  it  may  stand  for 
the  number  of  thousand  units  or  the  number  of  million  units  as  the 
problem  may  suggest. 

50.  In  1912  the  amount  of  gold  produced  in  Colorado  was 
1.1  millions  less  than  that  produced  in  Cahfornia;  the  amount 
produced  in  Alaska  was  1.4  millions  less  than  that  produced 
in  Colorado;  the  amount  from  Nevada  was  3.35  millions  less 
than  I  of  that  of  California;  the  amount  from  South  Dakota 
was  1.15  millions  more  than  J  of  that  from  Nevada;  from 
other  states  it  was  1.1  millions  less  than  2  times  that  from 
South  Dakota;  the  total  is  2.3  millions  less  than  5  times  that 
produced  by  Colorado.  What  was  the  amount  produced  in 
each  state  ?  What  was  the  total  gold  product  of  the  United 
States  for  that  year  ? 

51.  In  1910  the  number  of  people  Hving  in  rural  districts 
in  the  United  States  was  6.7  millions  more  than  the  number 
of  people  living  in  cities.  The  total  number  of  people  was 
91.9  milHons.  How  many  people  lived  in  the  rural  districts  ? 
How  many  in  the  cities  ? 

52.  In  1909  the  value  of  the  newspapers  and  periodicals 
in  this  country  was  placed  at  53  millions  less  than  twice  the 
value  in  1899.  The  value  in  1899  was  157  miUions  less  than 
the  value  in  1904.  The  total  value  for  the  three  years  was 
1684  millions.  What  was  the  value  placed  on  them  for 
each  of  the  three  years  ? 

53.  Class  should  look  up  statistics  and  write  three  prob- 
lems and  solve. 

To  form  Equations  Expressing  the  Relation  between  Un- 
known Numbers. 

Translate  the  English  into  algebraic  language,  then  form 
the  equation  and  reduce  it  by  means  of  the  axioms.  Thus 
we  can  find  the  relation  between  numbers  in  which  we  are 


x.§58]  SIMPLE  EQUATIONS  119 

interested,  even  though  we  do  not  know  the  values  of  the 
numbers. 

Example  1.     A  certain  sum  of  money  placed  at  simple  interest  at 
6%  amounts  in  a  year  to  the  same  as  another  sum  placed  at  interest 
at  8%.     How  does  the  first  sum  invested  compare  with  the  second? 
Solution: 

Let  p  =  the  number  of  dollars  invested  at  6% 

and  q  =  the  number  of  dollars  invested  at  8%. 

Then     l.OQp  =  the  number  of  dollars  in  the  amount  at  the  end  of 
year  of  money  at  6% 
and  1.08  g  =  the  number  of  dollars  in  amount  at  end  of  year  of 

money  at  8%; 
therefore    1.06  p  =  1.08  g,  by  statement  of  the  problem. 
1.08 

p  =  t:o6^ 

That  is,  the  capital  invested  at  6%  is  l^\  as  much  as  the  capital  invested 
at  8%.     We  have  not  found  the  value  of  either  p  or  q. 

Example  2.  Given  two  angles  that  are  equal,  to  find  the  relation  be- 
tween their  complements. 

Solution:  Since  the  angles  are  equal,  the  results  obtained  bj'  sub- 
tracting each  of  them  from  a  right  angle  are  again  equal.  (Axiom  III.) 
But  these  results  are  the  complements  of  the  given  angles;  therefore 
the  complements  of  equal  angles  are  also  equal  angles. 

Example  3.  If  two  lines  are  crossed  as  in  the  figure,  find  the  rela- 
tion between  angle  DEB  and  angle  CEA. 


Solution: 

Let  a  =  the  number  of  radians  in  angle  DEB, 

and  ai  =  the  number  of  radians  in  angle  CEA, 

and  h  =  the  number  of  radians  in  angle  BEC\ 

then  a  +  h  =Tr,  Why? 


120  ALGEBRA  —  FIRST  COURSE  [x.§58 

and  6  +  oi  =  tt;  Why? 

therefore        a  -\-  h  =  b  +  ai.     Why? 

Subtracting  h  from  each  member  of  the  equation, 

a  =  tti. 

That  is,  the  angle  DEB  is  equal  to  the  angle  CEA. 

In  the  figure  above  the  angles  DEB  and  CEA  are  called 
vertical  angles,  as  are  also  the  angles  EEC  and  AED. 

Exercise.  Let  the  student  solve  for  the  relation  between 
angles  EEC  and  AED. 

These  results  may  be  stated  thus: 

If  two  lines  intersect,  the  vertical  angles  are  equal. 

Exercise.  In  the  figure  of  Example  3,  draw  a  line  FG 
parallel  to  CD,  making  the  following  figure. 


By  measurement  show  that  angle  GME  is  equal  to  angle 
CEM,    Assume  that  this  is  true. 
Then  by  algebraic  solution  find  the  relation  between 

angle  MED  and  angle  EMF, 
angle  EEC  and  angle  AMG, 
angle  DEB  and  angle  FMA, 
angle  GME  and  angle  MED, 

Exercise.     In  the  last  figure  add  a  line  HK,  parallel  to 
AB,  forming  the  following  figure.    State  relations  between 


X.  §  581 


SIMPLE  EQUATIONS 


121 


the  various  angles.     What  can  you  say  about  the  angles  of 
the  parallelogram  ELNM? 


B 


15. 


Solviion  of  More  Difficult  Equations. 
Example.     Solve  the  following  equation  for  r: 

(3r-2)(|r  +  3)  -  (2r  +  1)  (2r  -  1)  =  |r» - 
Doing  the  work  indicated, 

[3r(fr  +  3)  -2(|r  +  3)]  -  [2r  (2r  -  1)  +  1  (2r  -  1)] 
[(4^r2  +  9r)  -  (3r  +  6)]  -[(ir^-2r)  +  (2r  -  1)] 
(4ir2  4-6r-6)  -  (4r2  -  1) 
§r2  +  6r-5 

Subtracting  ^  r^  from  both  sides  of  the  equation,  and  also  adding 
5  to  both  sides,  we  have 

6  r  =  -  10.     Why? 


= 

hr' 

-  15 

- 

\r' 

-  15 

= 

\r^ 

-15 

= 

hr' 

-  15 

Check  by  substituting  —  f  for  r  in  the  first  member  of  the  equa- 
tion. 

[3  (-i)  -  2]  [f  (-  f)  +  3]  -  [2  (-  f)  4- 11  [2  (-  ^)  -  11 

=  (-  5  -  2)  (-  f  +  3)  -  (-  Y  +  1)  (-  ¥  -  1) 

=  -7.^-(-i)(-Y) 

=  -!-¥ 

63  +  182 


18 


—  ¥/. 


122  ALGEBRA  —  FIRST  COURSE  tx.§69 

Substituting  —  |  for  r  in  the  second  member  of  the  equation, 

i(_5)2_15   =   1.2^_15 

=  ft  -  15 

=  --¥/-. 

Therefore  r  —  —  f  is  the  value  which  makes  the  equation  true. 

Exercises.*    Solve: 

1.  3  r  -  2  (2  -  r)  =  21. 

2.  3  (s  -  1)  =  4  {s  +  1). 

3.  (2  -  a)  (5  -  a)  =  a\ 

4.  9  ^  -  3  (5  i  -  6)  =  -  30. 

5.  2  (w  +  3)  -  3  (m  +  2)  =  0. 

6.  p  (p2  +  1)  =  p  (p2  _  1)  +  9. 

7.  3  a;  +  14  -  5  (x  -  3)  =  4  (a;  +  3). 

8.  w  {I  -[■  w  +  w^)  =  w^  -\-  w^  -\-  Zw  -  11 '  ^. 

9.  {x  +  1)  (a:  +  2)  -  (a;  +  3)  (x  +  4)  =  -  50. 

69.   Summary. 

To  solve  problems  leading  to  equations: 

Let  any  letter,  usually  the  initial  letter  of  the  name  of  the 
thing  whose  value  you  are  trying  to  find,  represent  the  num- 
ber expressing  that  value. 

Then  by  analysis  find  two  different  algebraic  expressions 
which  express  the  same  amount  of  something  spoken  of  in 
the  problem,  or  brought  out  indirectly  by  the  nature  of  the 
problem. 

Place  these  two  algebraic  expressions  equal  to  each  other 
in  the  form  of  an  equation. 

Solve  this  equation  by  means  of  the  axioms  —  bringing 
the  terms  which  contain  the  unknown  quantity  to  one  side 
of  the  equation  and  the  known  terms  to  the  other  side. 

Divide  both  sides  of  the  equation  by  the  coefficient  of  the 
unknown  number. 

*  Chapter  II  of  Geometry  may  be  taken  up  while  the  class  is  solving 
'the  exercises  and  problems  which  conclude  this  chapter. 


X.§591  SUMMARY  —  PROBLEMS  123 

Miscellaneous  Exercises. 

1.  A  man  earns  e  dollars  a  day  and  spends  r  dollars  a  day 
for  his  room  and  h  dollars  a  day  for  his  board.  If  he  works 
e  days  during  the  month,  how  much  money  has  he  at  the 
end  of  the  month,  if  his  board  and  room  rent  are  to  be  paid 
whether  he  works  or  not  ? 

2.  Write  a  problem  makmg  the  above  exercise  special. 
After  solving,  substitute  the  special  numbers  used  in  this 
exercise  in  the  answers  of  Exercise  1  and  compare  answers. 

3.  A  measuring  line  is  30  c  feet  and  2  h  inches  long.  When 
applied  to  a  line  it  goes  3J  c  times.  What  is  the  length  of 
the  line  measured  ? 

4.  Simplify  (a  +  6)  (a^  -\- 2  ah -\- h'')  -  2  ah  {a  ^  h). 

6.  A  man  agrees  to  work  for  5  dollars  a  day  for  every  day 
that  he  works,  during  which  time  he  pays  no  board  or  room 
rent.  For  the  days  that  he  does  not  work  he  must  pay  1 
dollar  a  day  for  board  and  room  rent.  If  he  works  20  days 
during  the  month,  how  much  money  should  he  receive  at 
the  end  of  the  month  ? 

6.  Make  a  general  problem  for  the  above  exercise.  Solve 
and  check  by  placing  the  special  values  used  in  Exercise  5 
in  the  answer  of  this  one.  Compare  with  the  answer  of 
Exercise  5. 

7.  Simplify  and  check  {m  -  7)  (m  +  9)  -  (w  +  3)  (m  -  2). 

8.  If  m  men  do  a  piece  of  work  in  t  days,  what  part  of  the 
work  will  one  man  do  in  one  day  ? 

9.  Make  the  above  problem  special,  and  check  as  in 
Exercises  1  and  2. 

10.  A  train  runs  at  the  rate  of  r  miles  an  hour,  stopping  s 
minutes  at  each  station.  If  it  takes  it  t  hours  to  make  the 
trip  and  it  stops  at  n  stations,  what  distance  has  it  traveled  ? 

11.  Make  the  above  exercise  special  and  use  as  check  for 
Exercise  10. 

12.  {w  -\-l){w-\-2)  -  {w-  3)  (w;  -  4)  =  0.  Solve  for  w 
and  check. 


124  ALGEBRA  —  FIRST  COURSE  [x.§59 

13.  A  man  earns  $5  a  day.  If  during  a  certain  month  he 
works  21 1  days  and  pays  75  cents  a  day  for  board,  50  cents 
a  day  for  rent,  and  75  cents  a  week  for  laundry,  how  much 
money  will  he  have  at  the  end  of  the  month  ? 

14.  Make  the  above  exercise  general  and  use  as  check. 

16.  A  man's  yearly  income  is  52  w  dollars;  his  expenses 
average  e  dollars  a  week.  He  enjoys  a  vacation  of  v  weeks, 
during  which  time  his  income  is  the  same,  but  his  expenses 
are  increased  by  20  dollars  a  week.  How  much  has  he  at  the 
end  of  the  year  ? 

16.  (^  -  3)  (2  ^  -  5)  -  2  ^  (^  -  8)  =  1.  Solve  for  the  value 
of  t. 

17.  A  cubic  centimeter  of  lead  and  a  cubic  centimeter  of 
copper  together  weigh  20.13  grams.  A  ball  containing  5  cc. 
of  copper  and  7  cc.  of  lead  weighs  27.07  grams  more  than  a 
ball  containing  4  cc.  of  lead  and  6  cc.  of  copper.  Find  the 
density  of  each  metal. 

18.  In  a  parallelogram  whose  adjacent  sides  are  a  and  6, 
if  10  units  were  added  to  the  side  a  and  3  units  subtracted 
from  the  side  h,  the  sides  of  the  figure  would  be  equal.  What 
is  the  length  of  each  side? 

19.  If  d  yards  of  cloth  cost  c  dollars  and  h  yards  are  sold 
for  a  dollars,  what  is  the  gain  on  each  yard  sold  ? 

20.  Make  Exercise  19  special.  Solve  and  show  corre- 
spondence of  answers. 

Momentum.  The  subject  that  we  shall  now  look  into  as 
to  its  measurement  is  the  amount  of  motion  in  a  moving 
body.  The  name  given  to  the  "amount  of  motion"  is  mo- 
mentum. It  is  measured  by  the  product  of  the  weight  by 
the  speed  of  the  moving  body. 

The  amount  of  motion  will  be  3  times  as  much  if  a  ball 
weighing  6  grams  moves  at  the  rate  of  100  cc.  per  second 
as  it  will  be  if  a  ball  weighing  2  grams  moves  at  the  rate  of 
100  cc.  per  second. 

Furthermore,  the  amount  of  motion  will  be  5  times  as 


X,  §  59]  SUMMARY  —  PROBLEMS  125 

much  if  a  body  weighing  6  grams  moves  at  the  rate  of  500  cc. 
per  second  as  it  will  be  if  a  body  weighing  6  grams  moves  at 
the  rate  of  100  cc.  per  second. 

Since  velocity  is  a  directed  quantity,  momentum  may  be 
regarded  as  a  directed  quantity. 

The  unit  of  measure  selected  for  momentum  is  the  amount 
of  motion  of  1  unit  of  matter  moving  with  a  velocity  of  1 
unit  of  distance  during  1  unit  of  time. 

In  the  centimeter-gram-second  system,  the  unit  of  mo- 
mentum is  the  momentum  of  1  gr.  of  matter  moving  at  the 
rate  of  1  cm.  per  second. 

21.  What  is  the  momentum  of  a  cannon  ball  weighing 
16  lbs.  moving  at  the  rate  of  250  ft.  per  second  ? 

In  this  exercise  let  the  momentum  of  1  lb.  moving  at  the 
rate  of  1  ft.  pe*-  second  be  the  unit  of  measure. 

Then  16  =  the  number  of  units  of  momentum  of  16  lbs. 
moving  at  the  rate  of  1  ft.  per  second 
and    4000  =  the  number  of  units  of  momentum  of  16  lbs. 
moving  at  the  rate  of  250  ft.  per  second. 

22.  Express  the  momentum  of  a  ball  weighing  w  grams 
and  moving  at  the  rate  of  v  cm.  per  second.  Let  M  stand 
for  momentum. 

23.  What  is  the  formula  for  the  expression  of  momentum? 

24.  Compare  the  momentum  of  a  street  car  weighing  5 
tons  going  at  the  rate  of  10  miles  an  hour  to  that  of  an 
engine  weighing  20  tons  and  going  at  the  rate  of  40  miles  an 
hour.  Solve  this  exercise,  then  make  it  general  and  solve. 
Compare  answers  by  substituting  the  special  values  given 
into  the  general  answer  obtained. 

26.  If  a  man  weighing  w  lbs.  is  riding  a  wheel  weighing  w 
lbs.  at  the  rate  of  v  ft.  per  second  due  east,  what  is  the 
momentum  ?  Would  the  momentum  be  the  same  if  he  were 
riding  south?  How  would  it  compare  if  he  were  going 
west? 

26.  A  body  composed  of  two  masses,  one  weighing  a 


126  ALGEBRA  —  FIRST  COURSE  [x,§59 

pounds  and  the  other  weighing  h  pounds,  is  moving  with  a 
velocity  of  {a  +  6)  ft.  per  second.  What  is  the  momentum 
of  the  body? 

27.  Two  balls  have  equal  momenta.  The  first  weighs 
100  kilos  and  moves  with  a  velocity  of  20  meters  per  second. 
The  other  moves  with  a  velocity  of  500  meters  per  second. 
How  much  does  it  weigh?  Solve  this  exercise  as  it  stands, 
then  make  it  general  and  solve.  Compare  answers  by  sub- 
stituting the  special  values  given  into  the  general  answer 
obtained. 

28.  The  velocity  of  a  body  is  2  y  meters  and  the  velocity 
of  another  is  7  meters.  If  each  weighs  (7  v  —  4)  grams,  how 
much  more  is  the  momentum  of  the  second  body  than  that 
of  the  first  ? 

29.  A  boy  weighs  50  pounds  more  than  his  bicycle.  When 
he  rides  at  the  rate  of  10  miles  an  hour,  the  momentum 
acquired  is  960  lbs.     Find  the  weight  of  the  boy. 


CHAPTER  XI 

SIMPLE  AREAS   AND   THEIR  ALGEBRAIC   EXPRES- 
SION.    ELEMENTS   OF   FACTORING 

60.  Rectangles.  In  the  chapter  on  measurement  (Chap- 
ter II),  we  brought  out  the  idea  that  to  measure  an  area  we 
usually  take  a  square  unit  and  find  out  how  many  such 
square  units  and  fractional  parts  of  such  square  units  are 
required  to  make  the  given  area. 

Definition.  A  four-sided  figure  all  of  whose  angles  are  right 
angles  is  called  a  rectangle. 

In  Chapter  II  we  found,  by  counting,  that  if  we  drew  a 
rectangle  with  a  definite  number  for  a  base,  and  a  definite 
number  for  its  height,  the  number  of  square  units  in  its  area 
is  equal  to  the  product  of  these  two  numbers.  Or,  express- 
ing this  in  algebraic  language : 

If  the  base  of  a  rectangle  consists  of  h  units  and  the  height 
of  h  units,  the  area  is 

b  '  h  =  bh  square  units. 

We  may  therefore  regard  the  product  of  any  two  positive 
numbers,  m,  n,  as  the  area  of  a  rectangle  whose  dimensions 
are  m  and  n. 

This  is  called  the  rectangle  m  •  n. 

A  square  whose  side  is  m  has  the  area  m  •  w  or  m^. 

Definition.  Two  rectangles  are  said  to  be  equivalent  if  they 
have  equal  areas. 

Exercise.  Given  that  a  rectangle  whose  dimensions  are 
a  •  6  is  equivalent  to  a  rectangle  whose  dimensions  are  c  •  d. 
Express  this  as  an  equation. 

127 


128 


ALGEBRA  — FIRST  COURSE 


IXI,  §  61 


61.  Geometric  Theorems  on  Factoring. 

Addition  of  rectangles.  As  in  Chapter  IX,  we  can  express 
multipHcation  of  polynomials  as  a  combination  of  rectangles. 

Draw  the  rectangle  whose  base  is  6  +  e  and  whose  height 
is  h  units.     Thus: 


Is  the  area  of  the  rectangle  h  (b  -\-  e)  equal  to  the  sum  of 
the  area  of  rectangles  hb  and  hel  Write  the  algebraic  equa- 
tion for  this. 

We  assume  here  and  in  what  follows  that  the  letters  stand 
for  positive  numbers. 

Do  similarly  with  m  {h  -{-  k  -\-  l) . 

We  can  state  this  as  a  theorem. 

Theorem  1.  The  rectangle  of  two  given  lines  is  equal  to  the 
sum  of  the  rectangles  of  one  of  them  and  the  several  segments 
into  which  the  other  is  divided. 

Exercises.  Draw  the  following  and  write  the  algebraic 
expression  for  the  equality  of  areas. 

1.  s  (2  s  +  r  +  t).  4.   am  +  an. 

2.  c{d  +  e  +  3c).  6.   bh  +  br  +  bd. 

3.  m  (a  +  b).  6.   a^  -}-  ab  -\-  ac. 

Express  the  following  algebraically  as  single  rectangles: 


7. 

bh  +  bk. 

13. 

5r2  +  10r. 

8. 

bh-{-bk  +  bl. 

14. 

2  m^  -\-  Q  mn  -\- 2  m. 

9. 

ah-{-bh-]-ch  +  dh. 

15. 

14s  +  28r.s  +  7r. 

10. 

m^  -\-  mn. 

16. 

15p'  +  5p  +  5. 

11. 

2h'  +  Shk. 

17. 

12  mn^  +  6  mn. 

12. 

c^  +  2c  +  3cd. 

18. 

iab^  +  Sah  +  l2a^ 

XI.  §61]        FACTORING  —  GEOMETRIC  THEOREMS  129 

Definition.     When  an  expression  like 
c2  +  2  c  +  3  cci 
is  rewritten  in  the  form  of  a  product, 

c  (c  +  2  +  3  d), 

this  operation  is  called  factoring. 

The  above  exercises  are  simple  exercises  in  factoring, 
illustrated  geometrically. 

Theorem  2.  The  square  on  the  sum  of  two  lines  equals  the 
sum  of  the  squares  on  those  lines  plus  twice  their  rectangle. 


a+h 

u 

M 

Q 

ab 

b 

R 

b' 

s 

T 

a 

b 

+ 

n^ 

a 

ab 

0 

P 

R 

i 

* 

a+h 

We  have  given  the  square  on  the  sum  of  the  two  lines  a 
and  h.     Call  it  OKTM. 

To  show  that  square  OX TM  is  equal  to  two  squares,  one 
of  which  has  a  side  a,  the  other  has  a  side  b,  and  two  rec- 
tangles of  the  dimensions  ahy  h.  In  other  words  we  are  to 
show  that 

(a  +  6)2  =  a2  +  2  aft  +  &*• 

Name  the  parts  of  which  the  square  OKTM  is  composed 
and  give  the  dimensions  of  each. 


130  ALGEBRA  — FIRST  COURSE  ixi.  §  61 

Make  the  algebraic  multiplication  of  (a  +  6)  (a  +  6).  Show 
the  correspondence  in  the  figure  of  each  partial  product. 

Exercises.  Make  a  geometric  representation  and  algebraic 
expression  of  the  following: 


1. 

(m+l)(m  +  l).                  5.   iit  +  s)(it  +  s). 

2. 

(2  m  +  n)  (2  w  +  n).            6.   c^  +  2  cd  +  d\ 

3. 

(r  +  3s)(r  +  3s).                7.   a'  +  2a+l. 

4. 

{2h-\-Sk){2h  +  Sk).         8.  m2  +  6m  +  9. 

9.   4s2  +  4rs  +  r2. 

Give  the  algebraic  expression  of  the  following  without 
drawing: 

10.  m^  +  4:m-\-  4.  12.   9  r^  +  12  rs  +  4  r^. 

11.  9  +  6/i  +  /i2.  13.   36  +  2-30  +  25. 

On  the  same  principle  as  above  give  the  geometric  picture 
of  the  following  rectangles,  give  the  algebraic  solution,  and 
point  out  the  geometric  representation  of  each  step  in  the 
solution. 


Exercises. 

1.  (r  +  2)(r  +  3). 

2.  (6 +  7)  (6  +  1). 

3.  (c  +  8)(c  +  5). 

4.  ia  +  b){a  +  c). 

6.    (p  +  r)(p  +  s). 

6.  c2  +  5c  +  6. 

7.  m2  +  7  m  +  10. 

8.  a^  +  ac-{-  ah  +  ch. 

Give  the  following  without  drawing: 

9.   s2  +  3s  +  2. 

10.  x^  +  7x+  12. 

11.  2/' +  82/ +12. 

12.  u'-\-lSu+12, 

13.  a^  +  as-^bs-{-  ah. 

14.  k^  +  2ak  +  Zhk  +  Qab, 

Subtraction  of  rectangles.  Draw  the  rectangle  h  (b  —  e). 
Do  this  by  drawing  length  b,  then  subtract  length  e,  e  being 
shorter  than  6.  Draw  height  h  and  complete  the  rectangle. 
Thus: 


xi.§6ij        FACTORING  —  GEOMETRIC  THEOREMS 


131 


The  rectangle  MNQP  is  the  rectangle  h  {h  —  e).  Rectan- 
gle ONQS  is  rectangle  hb;  OMPS  is  rectangle  he;  rectangle 
OMPS  minus  rectangle  ONQS  is  rectangle  MNQP. 


r 

( 

3 

he 

hb-he 

h 

e 

M 

be 

N 

«_ 

-b > 

1 

Therefore        hh  -  he  =  h  (b  -  e).     Why? 

Do  the  same  with  m  (h  —  k  +1);  that  is,  show  that  this 
is  equivalent  to  rectangle  mh,  minus  rectangle  mk,  plus 
rectangle  ml. 

Exercises.     Resolve  the  following  into  simpler  rectangles: 

1.   m  (a  -  6).        2.   5  (2  s  +  r  -  0.         3.   c  (d  -  e  +  3  c). 


Express  the  following  in  single  rectangles: 
4.   c'^-\-2c-Sdc.     5.   5r2-10r.    6.   2m^ 


6  mr  +  2  m. 


Theorem  3.  The  square  on  the  difference  between  two  lines 
equals  the  sum  of  the  squares  on  the  two  lines  minus  twice  the 
rectangle  of  the  lines. 

To  draw  the  square  whose  side  is  (a  —  b),  draw  length  a; 
from  it  subtract  length  b.  The  remainder  a  —  b  is  the  base. 
In  like  manner  draw  the  altitude  a  —  6,  so  as  to  form  a 
square.     (See  figure  on  next  page.) 

The  resulting  figure  PQRS  is  the  square  (a  —  6)  (a  —  6). 

OMRT  is  the  square  on  a.     Its  area  is  a\ 

OMQH  is  the  rectangle  ab. 

OKST  is  the  rectangle  ab. 

Rectangles  OMQH  and  OKST  are  both  subtracted  from 
square  OMRT. 


132 


ALGEBRA  —  FIRST  COURSE 


[XI.  §  61 


But  notice  that  after  subtracting  rectangle  OMQH  it  will 
be  necessary  to  add  square  OKPH  before  we  can  subtract 
rectangle  OKST,  so  that  rectangle  OMRT  minus  rectangle 


T 

.sr                               R 

H 
0 

(a-b)* 
P 

Q    ! 
M  ! 

b' 

b 

* 
b                                   b 

K              a-b 

a 

OMQH  plus  rectangle  OKPH  minus  rectangle  OKST  gives 
rectangle  PQRS. 

That  is:  (a  -  hy  =a^  -ah-\-¥-  ah, 

or,  (a  -hY  =  a^  -2ab^  b^ 

Exercises.     Draw  and  give  algebraic  expressions  for  the 
following: 

1.    {c-d)ic-d).  4.    (3r- l)(3r-l). 

•  2.    (m  -  2 n)  (m  -  2 n).  6.   h^  -2bc-\-  c^. 

3.    (3 -7)  (3 -7).  6.   4/i2_  4/^  +  1. 

Write  the  algebraic  expressions  for  the  following  without 
drawing;  in  other  words,  factor  the  given  expressions: 

7.  d^-2de-{-e\  9.   IG-Ss  +  s^. 

8.  9-6r  +  r2.  10.  25t^-d0t  +  9, 

11.  9m^n^  -  24m7i  +  16. 


XI.  §61]        FACTORING  —  GEOMETRIC  THEOREMS  133 

On  the  same  principle  as  above,  draw  and  give  algebraic 
expressions  for  the  following  rectangles. 

12.  (r  -  2)  (r  -  5).  15.    (a  -  6)  (a  -  c). 

13.  (b  -  7)  (6  -  1).  16.  m2  -  3  m  +  2. 

14.  (c-8)(c-5).  17.   p2_8p  +  15. 

Give  the  algebraic  expressions  for  the  following  without 
drawing: 

18.  r^  —  ar  —  br  -\-  ab.  20.   x^  —  14:xy  +  48  y^. 

19.  k^-2ak-Sbk-{-Q ab.     21.   d^  -15cd  +  5QcK 

22.   ^2  _  9  ^s  -I-  20  s\ 

Draw  the  following,  give  algebraic  expressions  for  the 
multiplication  and  show  correspondence  of  rectangles  to  the 
partial  products. 

23.  (a  -  5)  (a  +  3).  26.  {d-2e){d-{-  3/). 

24.  (c  +  7)  (c  -  2).  27.   m^-2m~  15. 

25.  (a  -  6)  (a  +  c).  28.   d^-}-2m-  15. 

Write  the  algebraic  expressions  for  the  dimensions  of  the 
following  rectangles  without  drawing. 

29.  c"^  —  7  cd  —  S  d^.  31.   p^  —  mp  -\- np  —  mn. 

30.  x^  -\-2xy  —  S  y^.  32.   k^  -\-  sk  —  rk  —  rs. 

33.   t^  -2rt  +  Sst-6  rs. 

Draw  the  following,  give  the  algebraic  expressions  for  the 
multiplication,  and  show  correspondence  of  rectangles  to 
the  partial  products: 

34.  (2  w  +  n)  (m  +  3  n).  36.    (2  m  -  n)  (m  +  3  n). 

35.  i2m  —  n)  {m  —  Sn).  37.   Qm^  -\- mn  —  2n^, 

38.   6  m^  —  mn  —  2  n^. 

Give  the  algebraic  expression  for  the  dimensions  of  the 
following  rectangles  without  drawing. 

39.  10r2  +  27rs  +  5s2.  41.   6a2+17a  +  3. 

40.  6a2-17a4-3.  42.  Sh^-lOh-S. 

43.   15  -  a;  -  6  x\ 


134 


ALGEBRA  —  FIRST  COURSE 


(XI.  §  61 


M 


Draw  the  following  and  state  dimensions: 

44.   an  —  hn-^ar  —  hr.  46.   2  pg'  +  2  ps  +  7  g  +  7  s, 

46.   ac  —  a  —  c  +  1. 
Without  drawing,  state  dimensions  of  the  following: 

47.  dy  —  cy  —  b  +  d. 

48.  r  (r  -  1)2  +  r  -  1. 

49.  (m  +  n){a-\-b)  -  (m-  n)  (a  +  6). 

50.  (s  +  2)(i-s)  +  (2-s)(^-s). 

Theorem  4.  The  difference  of  the  squares  on  two  lines  equals 
the  rectangle  of  the  sum  and  difference  of  the  two  lines. 

We  have  given  the  square  OKLM 
^  on  the  line  a,  and  the  square  OQRS 
on  the  line  b.    We  are  to  show  that 
the  difference  between  these  two 
squares   is  equal  to    a   rectangle 
<>  a  +  6  by  a  —  6. 

e      The  actual  difference  between 
these  two   squares   is,  by  defini- 
tion, the  irregular  figure  QKLMSR. 
Examine    the    rectangle    SRTM. 
What  is  the  length  of  its  side  SR  ? 
Of  its  side  SM?    If  we  take  this 
K  rectangle  and  place  it  with  its  side 
SM  coinciding  with  the  line  TL 
(why    do    they    coincide?),    thus 
subtracting    it    from    the    figure 
QKLMSR  and  adding  it  back  in  another  position,  we  shall 
have  the  rectangle  QKNP.     What  are  the  dimensions  of  this 
rectangle  ?    So  we  have 

a^  -b^  =  {a-\-  b)  {a  -  b). 

Exercises.  In  the  following  draw  the  difference  of  the 
squares  and  the  rectangle  to  which  this  difference  is  equal; 
write  the  algebraic  expression  for  this  equality. 


R 


b' 


a-h 


XI.  §61]        FACTORING  —  GEOMETRIC  THEOREMS  135 

1.  c^  —  4.  3.  m^  —  n^. 

2.  9-25.  4.   {a  +  b)^-c\ 

6.  m2  -  (p  +  q)\ 

Factor  the  following  without  drawing: 

6.  9  -  4  A2.  10.    (a  +  6)2  -  (a  -  6)2. 

7.  25  ^2  -  16.  11.    (m2  -  2  mn  +  n2)  -  4. 

8.  s2  -  1.  12.   9  -  (r2  4-  2  rs  +  s^). 

9.  1  -  4/2.  13.   52  _  y.2  _  2  rs  -  s2. 

Note.    When  it  is  necessary  to  inclose  terms  in  parentheses  after  a 
minus  sign,  the  sign  of  each  term  must  be  changed.    Explain  why. 

So  Exercise  13  would  first  be  written 


q^-  i^r-i-zrsi-s';. 

proceed  as  in  the  other  exercises. 

14. 

t' 

-  l+8p-  16p2. 

16. 

— 

r2  +  a2  -  2  a6  +  h\ 

16. 

49 

-m?  -2mn  +  n^. 

17. 

a2 

-  2  a6  +  fe2  -  r2  -  2  rs  - 

■s2. 

18. 

4- 

-  4  s  +  s2  -  n2  -  2  ng  +  ^ 

g2. 

19. 

1  - 

-  4  a2  +  8  a6  -  4  62. 

20. 

16 

-  8  6  +  62  -  a2  _|_  2  ac  - 

C2. 

21. 

a2 

+  2mn  +  ¥  -2ab  -m^ 

-n\ 

22. 

— 

4  r2  -  8  a6  +  a2  -  1  -  4  i 

r  +  16  h\ 

To  sum  up:  Two  numbers  which  multiplied  together  give 
a  certain  number  are  called  the  factors  of  that  number. 

The  process  of  finding  the  factors  of  an  expression  is  called 
factoring  the  expression. 

To  the  geometric  process  of  finding  the  dimensions  of  a 
rectangle  of  given  area,  there  corresponds  the  algebraic 
process  of  finding  the  two  numbers  which  multiplied  to- 
gether will  give  the  number  which  expresses  the  area.  In 
the  preceding  work  we  examined  both  processes. 

Factors  are  said  to  be  prime  when  they  cannot  be  re- 
factored. 


136  ALGEBRA  —  FIRST  COURSE  [xi.  §  62 

The  measurement  of  an  area  of  any  form  is  obtained 
essentially  through  the  measurement  of  an  equivalent  rec- 
tangular area.     We  shall  take  up  a  few  simple  cases. 

62.  Parallelograms. 

Definitions.  1.  A  parallelogram  is  a  four-sided  figure 
whose  opposite  sides  are  equal  and  parallel,  that  is,  the  same 
distance  apart  throughout  their  entire  length. 


/ 

\ 

/ 

\>^ 

/ 

N^ 

/ 

/ 

K 


2.  An  altitude  or  height  of  a  parallelogram  is  the  perpen- 
dicular distance  between  two  of  its  opposite  sides.  There 
are  two  altitudes,  Ai  and  /i2,  as  shown  in  the  figure. 

3.  When  the  altitude  is  drawn,  either  of  the  sides  to 
which  it  is  drawn  is  called  the  base  of  the  parallelogram. 

4.  A  parallelogram  with  equal  sides  is  called  a  rhombus. 
Experiment,     In  defining  a  parallelogram  we  said  that  the 

opposite  sides  are  equal  and  parallel.  Either  of  these  words 
may  be  omitted,  for  if  the  opposite  sides  are  equal,  they  will 
always  be  parallel;  if  parallel  they  will  always  be  equal. 
Without  proving  this  now  we  ask  the  student  to  verify  it 
experimentally. 

Cut  four  strips  of  cardboard  or  stiff  paper,  two  longer 
and  two  shorter  ones. 

Hinge  them  together  by  pins  or  thumbtacks  placed  at  the 
ends  of  the  lines  so  drawn  that  the  opposite  sides  are  equal, 
as  shown  in  the  figure. 


XI.  §  63] 


PARALLELOGRAMS 


137 


Swing  the  frame  into  various  positions.  See  if  the  opposite 
sides  are  always  parallel.  What  about  the  opposite  angles? 
What  have  we  shown  previously  that  the  sum  of  the  angles 


SX 


XJ 


of  a  quadrilateral  is  equal  to  ?  What  is  the  sum  of  any  two 
consecutive  angles  of  a  parallelogram  equal  to  ? 

How  does  the  area  change  as  the  figure  is  changed  from  a 
rectangle  to  a  more  oblique  form? 

63.  Theorem.  The  area  of  a  parallelogram  equals  the  area 
of  a  rectangle  of  the  same  base  and  altitude. 


Draw  the  parallelogram  ABCD,  with  base  6  and  height 
h.  Cut  out  the  figure.  Cut  off  the  triangle  APD  and 
place  it  in  the  position  of  triangle  BQC.  Does  it  coincide 
with  it  —  that  is,  does  it  fit  exactly  on  it  ?  A  rectangle 
is  formed  on  h  and  h  as  sides  and  with  the  same  area  as  the 
parallelogram  ABCD. 


138  ALGEBRA  —  FIRST  COURSE  [Xi,  §  64 

Corollary  1.     The  area  of  a  parallelogram  is  equal  to  the 
'product  of  its  base  by  its  altitude. 
That  is,  a  =  bh. 

Corollary  2.     All  parallelograms  having  equal  bases  and 
altitudes  are  equal  in  area. 

For  they  are  equivalent  to  the  same  rectangle. 


Exercises.  The  following  are  areas  of  parallelograms;  state 
the  lengths  of  their  bases  and  altitudes. 

1.  Kb  —  he. 

2.  c^  +  2  c  +  1.  What  kind  of  a  figure  is  this  parallelo- 
gram equal  to  ? 

3.  m2  -  4 n\     4.   r^  +  rs  -  pr  +  r.      5.   h^  +  3h-  4. 

64.  Triangles. 

Definitions.  The  three  lines  which  form  a  triangle  are 
called  the  sides  of  a  triangle.  The  three  points  of  intersection 
of  these  lines  are  the  vertices.  Each  is  called  a  vertex.  A 
line  drawn  from  a  vertex  perpendicular  to  the  opposite  side 
is  called  an  altitude  of  the  triangle.  There  are  three  alti- 
tudes. The  side  to  which  the  altitude  is  drawn  is  the  base 
of  the  triangle  corresponding  to  that  altitude. 

Theorem.  The  area  of  a  triangle  equals  half  the  area  of 
the  rectangle  having  the  same  base  and  altitude. 

In  our  measurement  work  in  Chapter  II  we  showed  this 
by  counting  the  number  of  square  units  in  each.  We  shall 
now  show  it  by  another  experiment. 

'    We  have  given  the  triangle  ABC  with  base  b  and  altitude  h. 
Form  a  parallelogram  by  drawing  CD  parallel  to  AB,  and 


XI.  §  64] 


TRIANGLES 


139 


BD  parallel  to  AC.  This  is  parallelogram  ABCD.  Cut 
out  the  figure  and  cut  the  paper  along  the  line  BC.  Fit 
triangles  one  on  the  other.     How  does  the  triangle  ABC 


compare  with  the  triangle  BDC?    Do  the  two  triangles 
coincide?    How  does  the  triangle  ABC  compare  with  the 
parallelogram   ABCD?    Therefore,  how   does  the  triangle 
ABC  compare  with  the  rectangle  on  h  and  h? 
The  algebraic  statement,  as  we  have  learned  before,  is 

a  =  ibh, 

where  a  stands  for  the  area  of  the  triangle. 

Corollary,    All  triangles  with  the  same  bases  and  altitudes 
are  equal  in  area. 


Explain  why  this  is  true. 

Exercise.  Take  three  strips  of  cardboard  all  of  different 
lengths.  Fasten  together  with  pins  as  in  the  experiment  with 
the  parallelogram.  Try  to  swing  the  figure  as  in  the  former 
case.     Wha,t  do  you  find? 

Exercises.  In  the  following  areas  of  triangles  tell  what 
may  be  the  length  of  the  base  and  altitude. 


6=?; 

A=  ?; 

6=?; 

/i=  ?; 

b  =  m  —  2  n; 

/i=  ?; 

h  =  m+l; 

h=  ?. 

140  ALGEBRA  —  FIRST  COURSE  p:i.§65 

Let  a  =  number  of  square  units  in  the  area;  b  =  number 
of  units  in  the  base;  h  =  number  of  units  in  the  altitude. 
a  =  24; 

a  =  m^  —  mn  —  2  n^; 
a  =  mn  —  rn  -\-  m  ~  r; 

65.  Regular  Polygons. 

Definitions.  A  polygon  is  a  figure  bounded  by  straight- 
line  segments. 

A  regular  polygon  is  one  whose  sides  are  all  equal  and  whose 
angles  are  all  equal. 

Construction  of  a  regular  polygon.  Draw  a  circle.  Divide 
the  circumference  into  any  number  of  equal  parts.  Join  the 
successive  points  of  division  by  straight  lines.  The  result- 
ing figure  is  a  regular  polygon. 

Plan  for  dividing  the  circumference  of  a  circle  into  equal  parts. 


Given  the  circumference  whose  center  is  O,  to  divide  it  into  five  equal 
parts. 

Draw  a  straight  line.  Cut  out  a  circle  of  same  size  as  circle  0. 
Mark  a  point  on  its  circumference.  Starting  at  this  point  roll  the 
circle  along  the  straight  line  until  the  point  again  comes  to  the  straight 
line.  You  now  have  the  line  segment  the  length  of  the  circumference. 
Call  the  line  segment  AB.  At  the  point  A  draw  a  line  of  indefinite 
length,  making  an  acute  angle  with  line  AB.  On  this  line  lay  off  a 
line  segment  of  any  convenient  length  five  times.  Call  the  last  point 
D.  Join  B  and  D  by  a  straight  Hne.  Then  at  each  of  the  other  points 
of  division  draw  an  angle  equal  to  angle  BDA.  Extend  the  arms  of 
these  angles  until  they  cut  line  AB,  and  AB  will  be  divided  into  five 
equal  parts. 

Now  roll  the  circle  again  marking  the  points  on  the  circumference. 


XI,  §66]  REGULAR  POLYGONS  141 

Definitions.  The  center  of  the  circle  is  the  center  of  the 
polygon. 

The  sum  of  the  sides  is  the  perimeter  of  the  polygon. 

The  line  from  the  center  perpendicular  to  one  of  the  sides 
is  called  the  apothem  of  the  polygon. 

66.  Theorem.  The  area  of  a  regular  polygon  equals  one- 
Jialf  the  product  of  the  apothem  by  the  perimeter. 

If  a  is  the  number  of  units  in  the  length  of  one  side,  n  the 
number  of  sides,  and  h  the  number  of  units  in  the  length  of 
the  apothem,  we  have 

na  =  the  number  of  units  in  the  length  of  the  perimeter. 
J  nah  •=  the  number  of  square  units  in  the  area  of  the 
polygon. 

Let  the  student  verify  this  by  dividing  the  polygon  into 
equal  triangles  by  lines  drawn  from  the  center  to  each  of 
the  vertices. 

Show  experimentally  that  these  triangles  are  all  equal  by 
drawing  a  regular  polygon  on  stiff  paper  and  cutting  it  up. 
Fit  the  triangles  on  one  another. 

Some  simple  regular  polygons  are: 

Three  sides:     equilateral  triangle. 

Four  sides:       a  square. 

Five  sides:       regular  pentagon. 

Six  sides:  regular  hexagon. 

Ten  sides:        regular  decagon. 
Exercises. 

1.  Draw  a  regular  pentagon  whose  center  call  0.  From 
0  draw  lines  to  each  of  the  vertices.  What  is  the  sum  of  the 
angles  in  one  triangle?  What  is  the  sum  of  the  angles  in 
the  five  triangles?  Since  the  angles  about  the  point  0  do 
not  make  up  the  angles  of  the  pentagon,  and  the  other  angles 
of  the  triangles  do  make  up  the  angles  of  the  pentagon,  if  we 
subtract  the  sum  of  the  angles  about  0  from  the  sum  of  the 


142 


ALGEBRA  —  FIRST  COURSE 


[XI,  §  67 


angles  of  the  triangles,  we  will  get  the  sum  of  the  angles  of 
the  polygon.  Doing  this,  what  do  we  get  for  the  sum  of 
the  angles  of  the  pentagon?  What  is  the  value  of  each 
angle  of  a  regular  pentagon  ? 

2.  Draw  a  regular  hexagon  and  use  it  in  the  same  way 
that  you  did  the  pentagon. 

3.  Draw  a  regular  decagon  and  examine  in  the  same 
manner. 

4.  ^  From  these  experiments  show  that  the  sum  of  the  angles 
of  these  polygons  may  be  expressed  by  (n—  2)  straight  angles, 
where  n  is  the  number  of  sides,  and  that  each  angle  may 


have  its  value  expressed  by 


straight  angles. 


5.  Calculate  the  value  of  each  angle  of  a  regular  seven- 
sided  figure,  assuming  that  the  above  expressions  always 
hold?     Of  a  nine-sided  figure?     Of  a  twenty-sided  figure? 

67.  Trapezoids. 

Definition,  A  four-sided  figure  with  two  of  its  sides 
parallel  is  called  a  trapezoid. 


IS 


jThe  two  parallel  sides  are  called  the  bases  of  the  trapezoid 

The  perpendicular   distance   between   the   two   bases 
called  the  altitude  (height)  of  the  trapezoid. 

Theorem.  The  area  of  a  trapezoid  is  equal  to  one-half  the 
product  of  its  altitude  hy  the  sum  of  its  parallel  sides. 

That  is,  letting  h  be  the  number  of  units  in  the  altitude, 
6i  the  number  of  units  in  the  lower  base  and  62  the  number 
of  units  in  its  upper  base, 


XI.  §  67] 


TRAPEZOIDS 


143 


i  /i  (&i  +  62)  =  the  number  of  units  in  the  area  of  the 
trapezoid. 

Let  us  have  given  the  trapezoid  ABCD, 


Draw  the  diagonal  AC.    This  gives  us  two  triangles  ABC 
and  ACD. 
So  we  have 

I  hhi  —  number  of  square  units  in  area  of  triangle 

ABC. 
§  hh2  =  number  of  square  imits  in  the  triangle  ACD. 
Adding: 
ihbi-\-  ^  hh2  =  number  of  square  units  in  the  area  of  the 
trapezoid. 
But 

ihhi  -\-  ihh2  =  ih  (hi  +  62),  by  factoring. 
This  shows  our  theorem  to  be  true. 

Exercises. 

1.  Construct  a  trapezoid  out  of  strips  of  cardboard  as 
was  done  in  the  parallelogram  in  §  62.  Can  you  swing  the 
figure  into  different  forms? 

2.  Draw  the  trapezoid  ABCD.  Let  E  be  the  mid- 
point of  side  AD,  and  F  be  the  mid-point  of  BC.  Draw  the 
line  EF.  Cut  out  the  trapezoid,  and  cut  it  in  two  along  the 
line  EF^  Place  the  two  parts  together  with  point  B  on 
point  C  and  BA  extending  in  a  line  with  DC.  What  is  the 
shape  of  the  figure  you  now  have?  Explain  how  this  may 
illustrate  our  theorem  about  the  area  of  a  trapezoid. 


144  ALGEBRA  —  FIRST  COURSE  [xi,  §  68 

3.  Suppose  that  we  have  given  that  the  altitude  of  a 
certain  trapezoid  is  2  units  and  the  sum  of  the  bases  is  10 
units.  Show  by  drawing  that  there  might  be  many  trape- 
zoids of  this  description.     Will  they  all  have  the  same  area? 

4.  Suppose  that  we  have  given  that  a  trapezoid  is  2  units 
high,  and  that  one  base  is  3  units  and  the  other  base  7  units, 
show  by  drawing  that  we  may  have  many  trapezoids  of  this 
description.     Are  they  all  of  the  same  area? 

68.   Circles. 

Definition.  A  circle  is  a  plane  figure  bounded  by  a  curved 
line,  all  points  of  which  are  equally  distant  from  a  point 
within  called  the  center. 

Circles  are  easily  drawn  by  the  aid  of  a  compass,  which 
the  student  should  secure.  The  kind  that  may  be  slipped 
on  an  ordinary  lead  pencil  will  do  very  well. 

There  is  some  confusion  with  regard  to  the  word  circle. 
Does  it  mean  the  boundary  line  or  does  it  mean  the  surface 
inclosed  by  this  line  ?  It  usually  means  the  surface  inclosed. 
The  boundary  line  is  called  the  perimeter  or  the  circumference 
of  the  circle.  * 

The  distance  from  the  center  to  the  circumference  is  called 
the  radius  of  the  circle. 

A  line  through  the  center  and  terminated  at  both  ends  by 
the  circumference  is  called  a  diameter  of  the  circle. 

The  diameter  is  always  twice  the  radius. 

The  circumference  of  a  circle  and  the  area  of  a  circle  may 
be  expressed  in  terms  of  the  radius  as  was  shown  in  §  16. 

At  that  time  we  showed  that 

c  =  27rr 
and  that  a  =  xr^. 

As  a  further  experiment  to  get  the  approximate  area  of 
the  circle,  draw  one  with  a  given  radius  on  cross-section 
paper.  Also  draw  a  square  whose  side  is  the  given  radius. 
Count  the  number  of  squares  in  the  area  of  the  circle,  adding 


XI,  §69]  FACTORING  CONTINUED  145 

together  the  parts  of  squares  to  make  the  whole  squares  as 
well  as  possible.  What  is  the  ratio  of  the  circle  to  the  area 
of  the  square?  Repeat  this  for  several  circles  of  different 
sizes,  thus  showing  that 


a  =  irr^. 

69.  Factoring  Continued.  In  the  first  part  of  this  chap- 
ter we  resolved  a  large  number  of  expressions  into  factors; 
we  also  noted  a  corresponding  geometric  process,  which 
consisted  in  expressing  the  combined  areas  of  several  rec- 
tangles as  a  single  rectangle. 

To  make  the  geometric  picture  of  the  process  we  supposed 
the  letters  contained  in  the  given  expressions  to  stand  for 
positive  numbers;  we  could  then  suppose  these  numbers  to 
represent  the  lengths  of  the  sides  of  rectangles.  But  the 
actual  process  is  just  as  good  when  the  letters  stand  for 
negative  numbers. 

For  example,  the  equation 

a2  -  62  =  (a  +  h)  (a  -  h) 

is  seen  to  be  true  by  direct  multiplication  of  the  right-hand 
member;  it  does  not  matter  whether  a  and  h  stand  for  posi- 
tive numbers  or  for  negative  numbers;  or  whether  h  is  less 
than  a  or  greater  than  a. 

In  this  article  we  merely  introduce  a  little  further  drill 
in  the  factoring  of  literal  expressions,  and  make  some  addi- 
tion to  our  stock  of  formulas  in  factoring. 

It  is  often  of  great  advantage  to  be  able  to  factor  expres- 
sions readily.  For  example,  suppose  that  it  is  required  to 
find  what  is  the  value  of  the  expression 

a^  +  Sa^  +  Sah^  +  h^, 

when  a  =  2  and  6  =  3. 
Substituting  we  find 

8  +  3. 4. 3  +  3. 2. 9  + 27  =  125. 


146  ALGEBRA  —  FIRST  COURSE  [xi.§69 

But  if  we  had  known  that  the  given  expression  is  equal  to 
(a  +  by,  that  is  (a  +  b)  {a -\-  b)  (a  +  b),  we  would  have 
had  its  value  thus: 

(2  +  3)3  =  5»  =  125. 

The  second  calculation  is  very  much  simpler. 
Again,  what  is  the  value  of  the  expression 

am  -{-bin  —  an  —  bn, 
when  a  =  3,  6=4,  w  =  8,  n  =  5? 

By  direct  substitution  we  have 

3. 8  +  4. 8-3. 5-4. 5=  21. 

But  the  given  expression  is  equal  to 

(a  +  6)  (m  —  n).' 

Substituting  in  this  we  have 

(3  +  4)  (8  -  5)  =  7  .  3  =  21. 

Factoring  is  especially  useful  when  an  expression  has  to  be 
calculated  for  several  values  of  the  letters  it  contains. 

Exercises.  Factor  the  following  expressions  and  calculate 
the  value  of  each  when  the  letters  are  replaced  by  the  given 
numbers.  Use  both  the  given  expression  and  the  factored 
form. 


1.   {a^-b^y, 

a  =  100,  b  =  99. 

2.  a2  +  2a6  +  62. 

a  =  5,  6  =  6. 

3.  a2  -  2  a6  +  6^; 

a  =  56,  6  =  53. 

4.  x2  +  2a;  +  l; 

x  =  2;3;5;  -  3;  -  5. 

6.  a;2-2a:+l; 

x  =  3;  6;  -5;    -7. 

6.  a;2  +  4a:  +  4; 

X  =  1;  3;  5;  -  1;  -3;  -5. 

7.   4m2  -  9n2; 

m  =  130,  n  =  86;  m  =  -  5, 

n  =-3. 

8.  u'^^-uv-2v^', 

1^  =  72,  t;  =  74. 

0.  4x2-  12a;l/  +  92/^• 

x  =  h  y  =  h 

XI.  §  69]  FACTORING  CONTINUED  147 

Other  useful  factors. 

Let  the  student  verify  by  direct  multiplication  the  fol- 
lowing equations;  then  memorize  them. 

(a  +  hy  =  a'-\-S  a%  +  3  06^  +  hK 
{a  -  6)3  =  a^  -  3  a^h  +  3  ofe^  -  h\ 
^3  +  6^  =  (a  +  6)  (a2  -  06  +  h^). 
a^-b'  =  {a-  h)  (a2  +  a6  +  b"^). 

State  these  rules  in  words. 

Example  1.     Let  a  =  2x  and  6  =  3?/;  then 

(2x  +  3i/)3  =  (2x)3  +  3  (2x)2  (3i/)  +  3  (2x)  (3!/)2  +  (3  y^ 
=  8  x3  +  36  x^i/  +  54  xy^  +  27  y^. 

Example  2. 

8x»-27i/3  =  (2x)3-  (3  2/)3 

=  (2x  -  3i/)  [(2x)2  +  {2x)  (3  2/)  +  (3y)2] 
=  (2x-32/)(4a;2  +  6xi/  +  9?/2). 


Exercises.     Write  the  e 

xpande 

dfo 

irms  of  the  foil 

1.    {a  +  2by. 

6. 

(3/i-4/b)3. 

2.    {a-2by. 

6. 

(3/i-4A;)3. 

3.    (2a;  +  2/)^ 

7. 

(5w  +  3v)3. 

4.    (3/i  +  4/c)3. 

8. 

(5z;-3t;)3. 

Factor  the  following: 

1.  x^-{-Sx^y  +  Sxy^  + 

2/3. 

6. 

64r3  +  125s3. 

2.   Sx'-3Qx^y+lSxy^- 

-27  2/3. 

7. 

a3+l. 

3.   8  w^  +  12  u^v  +  6  wy2 

+  z;3. 

8. 

l-a;3. 

4.   a'  +  S¥, 

9. 

1  -  xY' 

6.  m'  -  27  n\ 

Geometrical  picture  of  the  expansion  of  (a  +  6)3.  If  the 
dimensions  of  a  rectangular  block,  for  example  a  brick  or  a 
box,  are  a  =  number  of  units  in  length,  6  =  number  of  units 
in  width,  c  =  number  of  units  in  height, 

then,  a  •  6  •  c  =  number  of  cubic  units  in  the  contents 
or  volume, 


148  ALGEBRA  — FIRST  COURSE  IXI.§70 

For  example  a  box  4  ft.  long,  3  ft.  wide  and  2  ft.  high  con- 
tains 4  •  3  •  2  =24  cubic  feet. 

If  each  of  the  edges  of  the  block  has  a  length  a  units  its 
cubic  contents  are 

a  •  a  '  a  =  a^  =  the  number  of  cubic  units. 

Such  a  block  is  called  a  cube. 

Then  (a  +  6)^  =  the  number  of  cubic  units  in  a  cubical 
block  each  of  whose  edges  is  (a  +  6)  units  long.  (We 
suppose  a  and  h  to  stand  for  positive  numbers.) 

But  we  have 

(a  +  6)3  =  a^  +  3  a%  +  Sah^  +  h\ 

What  volume  is  represented  by  each  term  of  the  right- 
hand  member  of  the  equation. 

Make  a  paper  model  showing  each  of  these  volumes  and 
fit  them  together  so  as  to  form  the  cube  whose  edge  is 
(a  +  b). 

70.  Summary. 

Parallelograms  and  rectangles  are  equal 

If  they  have  the  same  or  equal  bases  and  lie  between  the 
same  parallel  lines; 

If  they  have  equal  bases  and  equal  altitudes. 

Triangles  are  equal 

If  they  have  the  same  or  equal  bases  and  lie  between  the 
same  parallel  lines; 

If  they  have  equal  bases  and  equal  altitudes. 

A  triangle  is  equal  to  one-half  of  a  parallelogram  having 
the  same  base  and  lying  between  the  same  parallel  Hues,  or 
having  the  same  base  and  altitude. 

The  area  of  a  rectangle  or  a  parallelogram  is  equal  to  the 
product  of  its  base  by  its  altitude. 

The  area  of  a  triangle  is  equal  to  one-half  of  the  product 
of  its  base  by  its  altitude. 


XI.  §  70]  SUMMARY  149 

The  following  equalities  have  been  shown. 
a2  +  2  a6  +  62  =  (a  +  by. 
a2  -  2  a6  +  6'  =  (a  -  hy. 
a^  -\-  (m  +  n)  a  +  mn  =  {a -{•  m)  (a  +  n), 
rsa^  +  (rm  +  sn)a-}-  mn  =  {sa  +  m)  {ra  +  n). 
a2  -  62  =  (a  _  h)  (a  +  6). 
a3  +  63  =  (ct  _|.  5)  (0^2  _  a5  +  52). 

a3  -  63  =  (a  -  6)  {a^  +  a6  +  62). 
(a  +  6)3  =  a^  +  3a26  +  3a62  +  63. 
(a  -  6)3  =  a3  -  3a26  +  3a62  -  6^ 


CHAPTER  XII 

FRACTIONS 

71.  Reduction  of  Fractions  to  Lowest  Terms  —  Cancel- 
lation. In  arithmetic  you  learned  how  to  reduce  fractions 
by  cancellation,  and  how  to  multiply  and  divide,  add  atid 
subtract  fractions.  In  algebra  these  operations  are  carried 
on  just  the  same  as  in  arithmetic;  we  shall  give  examples 
of  these  processes,  first  using  arithmetic  fractions  and  then 
literal  fractions. 

1575 
Example  1.    Simplify:   2520' 

5 

^  ,  .  1575      ^'-^^'-r , 

Factoring:  __  =  ^f_^_y 

5 
Cancelling:  =^- 

Why  does  cancelling  factors  from  the  numerator  and  the  denomi- 
nator of  a  fraction  not  change  its  value? 

q2  _  52 
Example  2.    Simplify:  3  ,  ,  3  • 

a2  _  62  (zH^  (a  -  h) 


Factoring: 
Cancelling: 


+  63      (cU^  (a'  -  a6  +  62) 
a-h 


a^-ab-h  62 


ExampUS.    Simpbfy:    60m2g2  -  6mng2  -  l26n2g2• 
12m2g  -  27 n^q  ^    Xr(2ja^-^n)  (2m  +  3n)  . 
i^actormg:  60m2g2-6wwg2_i26n232     2 -jS^iZm^-^^n)  {b m  + 1  n) 

r.        11-  2m  +  3n 

^^^^^^f^'  ==  2g(57n  +  7n)* 

160 


XII.  §  731  FRACTIONS  151 

72.   Multiplication  of  Fractions. 

„  ,    ,      cj.      ,.-        3     65     98 

Example!.    Simplify:    Jq  *  li  "  105* 


Factoring: 
Cancelling: 

Example  2. 
^^J^toring: 


J_    65    ^^  j;_   0'lS     %'p 
10*  14*  105     2-5*  ;2./*^.^yr' 

10* 

62  +  36  62-4c2      62 -26  + 26c -4c 


62  +  46c  +  4c2    62  +  6-6       262  +  6c-10c2 


(M^c)(M^^)*    (M<^)  (6.^=-^)    '(6/-2^)(2  6  +  5c) 


Cancelling : 


26  +  5C 


73.  Division  of  Fractions.  In  arithmetic,  to  divide  a 
number  by  a  fraction  is  the  same  as  to  multiply  the  number 
by  the  fraction  inverted. 

2  3 

Example  1.  6-^^  =  6•x=9. 

Go  back  to  the  definition  of  division  and  explain  why  this  is  so. 

This  division  could  have  been  performed  by  multiplying  the  divi- 
dend and  divisor  by  a  number  that  would  make  the  divisor  an  integral 
number  and  then  dividing.  Thus:  Multiplying  numerator  and  de- 
nominator by  3, 

6    3_  3»^»3  _ 
.2*3"      %       ~  ^• 


Example  2. 

Factoring: 
Cancelling: 


3 

63      a2  -  2  a6  +  62 


a +  6     '02  +  206  +  62 

^0^-63    a2  +  2a6  +  62 
a +  6  *a2  -  2a6  +62- 

Cg^-^(a2  +  a6  +  62)    (a-+i?;(a  +  6) 
(a-K6)  '  {q.^o){a,-h) 

_  (g  +  6)  (a2  +  a6  +  62) 
(a  -  &) 


152  ALGEBRA  —  FIRST  COURSE  fxii,§74 

Performing  this  division  by  the  second  plan  suggested  above: 
q3-63  (g  -  h)  (gz  +  ab  +  b^) 

^^"*^"°S:  a2-2ab  +  b^  =         {a  -  b)  {a  -  b)— 

a^  +  2ab  +  b^  {_a  +  b){a  +  b) 

Multiplying  both  terms  by  (a  +  6)  (a  +  6) : 

^  {a^^  (a  +  b)  (gg  +  ab  +  b^) 
{ff^-^o)  (a-b) 
Cancelling:  =  (a  +  5)  (gHh  g6  +  b^ 

Exercises. 

Simplify  the  following.     Check. 

Srh^    llkmV^    21  pV^  3  cd    2cd^ 

ma-\-mh  ^  m^  —  n^  aq  —  ap    p^-\-  pq 

a^  —  h^'    m^  —  2  mn  +  n^         *         p         p'^  —  q^ 


8. 


a  +  2  a2  +  a-6     a^  -a-Q 

-4s+4s2^  r-2s 


r3  +  8s3  r2-2rs  +  4s2 

74.  Reduction  to  a  Common  Denominator.  In  arithme- 
tic we  learn  that  to  add  or  subtract  fractions  requires  that 
they  be  expressed  in  the  same  denominator. 

For  example: 

2  4^  10      12  _22 

3  "^5      15  "^15      15* 

Now  any  set  of  fractions  can  be  reduced  to  a  conmion 
denominator  by  reducing  each  to  a  fraction  whose  denomi- 
nator is  the  product  of  the  denominators  of  the  several 
denominators;  but  this  is  often  not  the  lowest  common 
denominator. 

2     5      3 

For  example:  x,  -pz,  ^  may  all  be  reduced  to  fractions 

having  a  common  denominator  9  •  12  •  20  ==  2160. 


:,  §  74] 

FRACTIONS 

153 

2      2  .  12  .  20 . 
^^     9      9.12.20' 

5        5  .  9  .  20 
12      12.  9.  20' 

3        3.9.12 
20      20  •  9  .  12 

But  if  we  write  the  given  fractions  with  their  denominators 
in  their  factored  form,  we  have 

2  5  3 

3.3'     3.4'     4.5' 

Therefore  the  number  3.3.4.5  =  180  will  contain  all  the 
given  denominators,  and  this  is  the  smallest  such  number. 
It  is  called  the  least  common  denominator.  (We  shall  abbre- 
viate this  to  L.  C.  D.)     We  would  now  have 

2  ^  2  .  20 .        5  ^5.15^        3   ^3.9 
9  ~    180  '       12        180  '       20  ~  180  * 

It  is  better,  however,  not  to  multiply  out  the  factors  of  the 
denominator  until  you  have  added  the  numerators,  for  it 
only  makes  double  work  if  the  same  factor  is  found  in  the 
numerator  as  in  the  denominator.  You  would  first  multiply 
it  into  the  other  factors  and  then  divide  it  out  again.  A 
good  rule  to  follow  here  and  elsewhere  is:  Never  multiply 
out  until  you  have  to.  It  is  also  shorter  to  write  the  denomi- 
nator but  once  and  write  the  numerators  with  the  signs  of 
operation  above  it. 

rp.  2,5        3       40  +  75  -  27 

Thus:  -  +  _-_  = 


3.3.4.5 

88 

3.3.4.5 

^.22 

3.3.^4.5 

22 
45* 

Literal  fractions  are  handled  in  exactly  the  same  way. 


154  ALGEBRA  — FIRST  COURSE  [Xii,  §  75 

75.  Addition  and  Subtraction  of  Fractions. 

Rule.  Reduce  the  given  fractions  to  their  least  common 
denominator;  then  combine  the  numerators  with  their  proper 
sign. 

Example  1.  10'^75~3'^ 

Factoring  denominators  to  find  L.  CD.: 

9    .   14      2        9     .     14        2 


10*^75      3      2- 5"^  3-52      3 


Reducing  to  L.  C.  D.: 
Collecting: 
Factoring: 
Cancelling: 


135  +  28  -  100 
2  .  3  .  52       ' 
63 
2.3.52' 
3.21 


2-3-25 
21 
50' 
h    ,    a         c 


Example  2.  a'  +  b^c'^b^'' 

The  least  common  denominator  must  contain  a^,  h^,  and  c  as  factors; 
hence  it  is  a%^c.    Then,  reducing  to  L.  C.  D., 

^   ,   _a _c h-b^c        g.  a^  _    C'  dbc 

a'      h^c      a%  ~  o3  •  b^c  "*"  62c .  a^      a%  -  abc 


_  ¥c  +  a^  -  obc2 
Example  3. 


a^b^c 
2mn  2  m 


Factoring  denominator: 
Reducing  to  L.  C.  D.: 


m^  —  n^      m  -\-n 
n  2  mn  2  m 


m  —  n      {m  —  n)  (m  -\- n)      m  +  n 
n(m  -\-n)—  2 mn  —  2m(m  —  n) 


(m  —  n)  (m  +  n) 
Multiplying  terms  of  numerator: 

_  (mn  +  n^)  —2  mn  —  {2m^  —  2  mn) 
(m  —  n)  (m  -{■  n) 
n2  +  mn  —  2m^ 


Collecting  terms: 
Factoring: 


(m  —  n)  (m-{-  n) 
(n  —  m)  (n  +  2  m) 
(w  —  n)  (m  +  n) 


XII,  §  761  FRACTIONS  155 

ri        IT                                   —  (n  +  2  m) 
Cancelling:  =  — ^ — • 

We  might  have  written  the  answer  in  the  form ; —  .     Why? 

Is  Zll?=  J^=_2?    Why? 
o  —  o 

76.  Complex  Fractions.  A  complex  fraction  is  nothing 
more  nor  less  than  an  indicated  division  in  which  either  one 
or  both  of  the  numbers  involved  are  fractions  or  mixed 
numbers.  As  has  been  stated  the  easiest  way  to  handle  it 
is  to  reduce  to  a  simple  fraction  by  multiplying  the  numera- 
tor and  the  denominator  by  the  least  common  denominator 
of  all  the  fractions,  and  then  to  deal  with  the  simple  fraction 
as  you  have  been  shown  in  the  preceding  examples. 

Example  1.  ^  =  "^ 

Here  the  L.  C.  D.  of  numerator  and  denominator  is  14. 
Multiplying  both  terms  of        2^  _  33 
the  fraction  by  14:  6|  ~  88  ' 

Factormg:  =  g^-^- 


Cancelling: 
Example  2. 


^3 

8* 

f  +  1 


c  _bc+a^ 


b  52 

The  L.  C.  D.  of  numerator  and  denominator  is  h^. 

-  +  1 
Multiplying  both  terms  b _  db  +h^ 

by  L.  C.  D.,  namely  b^:.    ,c  _  be  +  a^  ~  b^  +  bc-  {be  +  a^) 
Collecting  and  factoring:  —  \     >     )       ^ 


Cancelling: 


(6  -  a)  QiA-a) 
b 


b  —  a 


Check  this  by  putting  numbers  for  a,  b  and  c  in  the  given  fraction 
and  in  the  answer;  for  example,  put  o  =  4,  &=—  2,  c  =  3. 


156  ALGEBRA  — FIRST  COURSE  [xil.  §  77 

Exercises. 

Simplify   the   following   and   check   exercises   containing 
letters. 


1. 

45      15^12 

m-\-  n      m  —  n 
m              n 

2. 

6^a' 

ac       he 
hd^      ad 

a-\-h         4:  ah 
a-h      a2-62 

3. 

2m 
Tin?-  —  7 

n 

.         r-4              r+6 
r2-5r  +  6  '  r2-6r+9 

^2                ^2     _     jy^^ 

7. 

a 

6                    6 

2  a2  +  5  afe  - 

■  3  62      2  a2  _  a6      a2  +  3  a6 

8. 

3a 

a^  -  b^      a^ 

+  a6  +  62  '  a2-a6 

9. 

2f  .    4  +  1, 

.    2i  +  3i 
'    21 -3i 

10. 

,   1      3m2 

—  4  mn  +  n2 
m2  —  ^2 

r 
^-6 

m  —  n 

m  +  n 

11. 

26r 

7r2+13r-2 

a2      a6  ^  62 

,      6r  +  l 
7r-l 

77.  Long  Division.  If  numbers  can  be  factored,  the 
easiest  way  to  divide  one  by  the  other  is  to  resolve  them  into 
their  factors,  and  then  cancel  the  common  factors.  How- 
ever, at  times  it  is  necessary  to  divide  one  number  by  another, 
when  neither  can  be  factored  by  the  rules  given.  In  this 
case  we  must  resort  to  the  use  of  long  division.  Long  di- 
vision in  algebra  is  similar  to  that  in  arithmetic. 


XII.  §77]  LONG  DIVISION  157 

Example  1.  Divide  1502115  by  6285. 

1502115   16285 
12570      239 

24511 

18855 


56565 
56565 


So  with  algebraic  expressions. 

Example  2.  Divide  a^  -  4  a^b  +  4  a%'^  -  ah*  by  a^  -  2  ah'+  h\ 
a^  -  4 a<6  +  4 a%^  -  a¥  \a^  -2ab  +  b^  (divisor) 
a'  -2a*b+  a^b^  a^  -  2  a%  -  ab^ '    (quotient) 

-  2  a^b  +  3  a^b^  +  0 
-2a*b  +  ia^b^-2a%^ 

-a%^  +  2a^b^  -  ab* 
-a%^  +  2a%^-abK 

Therefore    ^ ^    ,    ,   ,„ =  a^  —  2a%  —  ab^. 

a^  —  2ab  -\-b^ 

Check  this  by  putting  a  =  2  and  6  =  —  3. 

Rule  for  dividing  one  algebraic  expression  by  another. 

Arrange  the  terms  of  the  dividend  and  the  divisor  according  to  the 
exponent  of  some  letter  (a  in  our  illustration). 

Divide  the  first  term  of  the  dividend  by  the  first  term  of  the  divisor. 
The  result  is  the  first  term  of  the  quotient. 

Multiply  the  entire  divisor  by  this  term  and  subtract. 

Divide  the  first  term  of  the  remainder  by  the  first  of  the  divisor. 
This  gives  the  second  term  of  the  quotient. 

Multiply  the  entire  divisor  by  this  term  of  the  quotient  and  subtract. 

Continue  until  there  is  no  remainder  or  until  the  exponent  of  the 
leading  letter  of  the  remainder  is  less  than  that  of  the  divisor.  In  the 
latter  case  you  do  not  have  an  even  division.  Your  answer  then  is 
the  quotient  so  far  found  plus  a  fraction  whose  numerator  is  the  last 
remainder  and  whose  denominator  is  the  divisor. 

Check  by  multiplying  the  quotient  by  the  divisor.     If  the  work  has 
been  correctly  done,  the  result  is  the  dividend. 
6x4 -9x3 +  11  a:2- 10 


Example  3. 


3x2-2 
6x4-9x3  +  11x2-10     13x2-2  (divisor) 

6x4 -4x2  2 x2  -  3 X  +  5'    (quotient) 

-9x3  +  15x2-  10 

-9x3+    6x 

15x2  -  6x  -  10 
15  x2  -  10 


—  Q'z  (remainder). 


158  ALGEBRA  — FIRST  COURSE  ixii.  §  78 

Therefore   "^^  -  ^^l^.^  -  ^"  =  2.^  -  3.  +  5  +  3-^^  . 

Check  this  by  putting  a  number  in  place  of  x,  say  x  =  —  2. 

The  use  of  long  division  is  convenient  in  reducing  a  fraction  to  its 
lowest  term,  where  either  the  numerator  or  the  denominator  can  be 
factored  but  not  both.  The  plan  is  to  factor  the  term  of  the  fraction 
that  you  can  and  then  divide  the  other  term  by  one  of  the  factors.  If 
it  divides  without  a  remainder,  you  can  easily  write  the  fraction  in  its 
lowest  terms. 

q3  _  53 
Example  4.     Simphfy: 


Factoring: 


a*  +  a262  +  6* 
o3  _  53  (^a-  b)  (a2  +  ab  +  b^) 


a^  +  a262  +  54  a*^  a%^  +  ¥ 

Divide  a^  +  a^b^  +  ¥  by  a^  +  06  +  b^  by  long  division,  and  we 
have  the  quotient  a"^  —  ab  -\-  b^.     Therefore 

a'  -b^       ^        {a-b)(a^  +  ab  +  b^) 
a*  +  ab  +  62       {a^  +  ab  +  b^)  (a^  -  ab  +  b^) 
_        a  —  b 
~  a2  -  06  +  52  ■ 

Exercises.     Reduce  the  following,  and  in  each  case  check 
by  putting  numbers  for  the  letters. 

Divide: 

1.   x^  -]-  y^  by  X  -{-  y.  2.   x^  -\-  y^  hy  x  -\-  y, 

3.  r^  —  s^  by  r  —  s. 

4.  m^  +  4  m^n^  +  16  n^  by  m^  -  2  mn  +  4  n^. 
6.   p3  +  3 p^q  -  4:pq^  +  6  by  p^  -  pq  +  2 q. 

Reduce  the  following  fractions  to  lowest  terms: 

a^ -3x^ -13x  +  15  „    m4  +  ^2_}_i 

6.    ^ 7^ r-. 7. 


8. 


2  0^+1  m^- 

4  g^  -  9  b^ 

4a3-  20a26  -9a62  +  45  63" 


78.  Equations  involving  Fractions.  Fractions  often  occur 
in  equations  drawn  from  problems,  and  we  may  as  well  ex- 
amine at  this  time  the  method  of  dealing  with  such  equations. 


XII.  §78]        EQUATIONS  INVOLVING  FRACTIONS  159 

Suppose  that  we  had  formed  such  an  equation  as  the 
following: 

2t  +  l       2t  +  l 

(1  +  2)'^  t +  2  ""^ 

where  t  is  the  unknown  number. 

Since  this  is  an  equation,  we  may  multiply  each  term  in 
it  by  the  same  number  and  not  destroy  the  equality.  Why  ? 
The  values  of  the  individual  terms  will  be  changed,  but  we 
do  not  care  about  this  fact,  so  long  as  we  have  an  equation. 
So,  to  get  rid  of  fractions,  we  shall  multiply  each  term  of 
the  equation  by  the  number  that  will  multiply  out  the 
denominators.  The  best  number  to  select  is  the  least 
common  denominator. 

In  our  equation  the  least  common  denominator  is 

{t  +  2){t  +  2). 

Multiplying  each  term  of  the  equation  by  this  number,  we 
get 

(2^  +  1) +  (2^24.  5^ +  2)  -{2t^  +  St  +  S)  =0. 

Collecting  terms :  —^  —  5=0. 

Therefore  ^  =  —  5. 

Check  the  exercise  by  substituting  this  value  for  t  in  the 
equation.     We  have 

_  Q        _  Q 

'    -^  +  -::^-2=-i  +  3-2  =  o. 

Since  the  equation  states  that  the  value  should  be  zero, 
—  5  is  the  correct  value  of  t. 

Exercises.  J 

Solve  the  following  equations  and  check. 
2  m 


1. 


m-1      m2  +  3n-4 


p^  —  2p      p2_4p_|_4 


160  ALGEBRA  —  FIRST  COURSE  [xil.  §  78 

3.  1  '  ' 


6. 


terms : 

1.  ^;    ^ 

2. 


.2r2-9r  +  10 

^2  _  4      2  r2  -  r  -  10 

2+^-1    - 
^3a-2 

7a2 

3  a2  -  5  a  +  2 

m  —  1 
m^  -  3  m  " 

5m2  -  m  —  41 
m2-9 

r                 2 

'          +      ^       -1 

r  -  2      r2  -  4 

s  in  Fractions. 

Reduce  the  following  to  lowest 

4      15      64 
9'    25^    56' 

144      1442 
216'    3605 

a62     2^2      - 

- 12  m^ns^      51  a%'^cd 

ac'    6rfc' 

-Qmbh   '     d4:a%'d 

r2-9 

ah  +  h\    r^-\-2rh  +  rs^ 

r2-5r  +  6' 

ah+a^'        2r^  +  2rh 

2  p%  -  4  p2s2 

+  2ps\    a^-Sd' 

2  p2  _  8  ps 

+  6  §2     '    a^  -4:d^ 

3-3s3 

4  -  (a  +  6)2 

3  +  3  s  +  3  s2 

'    6-3a-36 

a2-6^    2r- 

-rs'—2s  +  s^ 

^'   h^  -a^'  7r-7s 

Multiply  the  following  and  see  that  answers  are  given  in 
their  lowest  terms: 


7. 

9. 
10. 
11. 


2 
5 

35    225 
15     14 

8.   ^ 

8  ms 
3s 

9s2 

2m 

P' 

■  -  2  p  +  1    P^ 

^+2p2  +  i 

p'-^P 

P- 

1 

r         1-s^ 

r 

-1 

s  —  1    mr  -\-  m   r  -}-  rs  -\-  rs^ 
r  (r  —  1)2  4-  r  —  1    y^  +  y 
dy  —  by  —  b  +  d    r^  +  1 


XII.  §78]        EQUATIONS  INVOLVING  FRACTIONS  161 

6a2-  lla  +  3    2a'-Sa^ 


12. 
13. 


6  a^  9  a* 

m       1  —  m^       m? 


m      —m      m^  + 1 

)2'(-a)3*a* 

(1  -  n)2       n2  -  1  2  -  n 


{-aY' {-ay' a  "'   W     W     3^« 

16. 


1+  n      2  +  n  -  n2    (n  -  1)^ 

Write  the  following  indicated  divisions  in  simplest  form: 

1  a      r       g^         m}_ 

.      ___,„,             2  6c     s       St        2n2 

17.    .,    ^  ,   ^,  ,    .  ,    .       ^  ^^'    d'    4t'    j_'       3 

6  c             3  s^     6  mn 

19.    7 21.    -: 7-  23 

b  —  a 


3 

7 

1' 

2 

8 

21.    li.   3K 

2  '   2i '     1  ' 

3  15 

3 

r 

4 

,   1 

1 

21. 

1       1 

b      a 

b^  a 

r2  —  2  rs  +  s 

22. 

r2  +  2  rs  +  s 

r  —  s 

a 
ab  r^  —  2rs  -{-  s  m^  —  n^  —  1      .. 

^"^a-6  r^  +  2  rs  +  s  2n     . 

a6  *  r  —  s  *        ,  n'^  —  m^  + 1 

^"^^^^  M^  "^         2n 


r  r 


„^    r  +  3      r-3     ^^ 
25. 26. 


r-3     r+3 
Perform  the  indicated  operation  in  the  following: 

27.    1+i;    3-i;    -7  +  2^;    -^-li- 
as    l-?  +  §.    2  ,  1_27. 
2      3^7'    3^12      72 
„     ,       1  ,1  ,  n     a      c 

^-  ^-;i'  '■+1;  -"^+2;  6-5- 

30.    -ll--l±I.  31.    a- ^i-. 

r  —  Ir  —  1  a  —  c 


162                       ALGEBRA  —  FIRST  COURSE  [xii.  §  78 

32.    FT-  +  1 -^ 


34. 


2r      4r      6r  m  —  1        w^ 

1      _     3  mn  m  —  n 

n  —  m      n^  —  w?      m^  +  mn  +  ^^ 


gg  5x 2x-^y 4a;  -y 

9x2  — 25 1/2      6nx  +  10ni/      6na;  — lOni/ 


62  _  (j5  '  a"  -  ob 
Solve  the  following  equations: 

37.  -^ —  q  ~  fi  ^  ^"    ^^l'^^  ^^^  value  of  r. 

38.  TT -. —  =  7^ .     Solve  for  value  of  s. 

5         4         20 

3  5  2 

39.    r. 7—^  =  —0 — 7,  •     Solve  for  value  of  m. 

m  —  6      m  +  3      m^  —  9 

40. 5 — Pi—  = ^.     Solve  for  value  of  x, 

X         x^  —  2x      X  —  2 

2rc  +  l 

3  3 

41. 7  =  TT.    Solve  for  value  of  x. 

Sx  —  1      2 


5  + 


6 
12  8 


42.  7-  =  — ri — .    Solve  for  value  of  r. 

7_4        _J_ 
r        2r  +  3 

Problems  leading  to  Simple  Fractional  Equations. 

(a)  Formation  of  the  Equation. 

(1)  Read  the  problem  with  care  entirely  through.  If 
there  is  anything  not  clearly  understood,  read  it  through 
again.  Be  sure  that  you  understand  the  English  statement 
before  you  begin  translating  into  algebraic  language. 


XII,  §78]        EQUATIONS  INVOLVING  FRACTIONS  163 

(2)  Decide  from  the  nature  of  the  problem  what  number 
is  described  in  two  different  ways.  This  is  very  important 
for  upon  this  depends  your  abihty  to  form  an  equation, 
which  is  of  the  utmost  importance  to  the  solution. 

(3)  If  there  is  but  one  value  called  for  by  the  nature  of 
the  question  let  a  letter  stand  for  this  value.  If  there  are 
several  numbers  whose  values  are  called  for,  let  a  letter 
stand  for  one  of  them,  and  form  algebraic  expressions  to 
stand  for  the  values  of  the  others;  these  algebraic  expressions 
are  merely  the  translations  of  the  relation  existing  among  the 
numbers  called  for,  as  expressed  by  the  English  of  the  problem. 

(4)  Translate,  clause  by  clause,  as  nearly  as  possible  in 
the  order  given  in  the  problem,  the  meaning  of  the  English 
language  into  algebraic  language.  You  will  always  get  two 
algebraic  expressions  which  stand  for  the  same  number. 
This  is  the  number  that  you  have  decided  upon  in  instruc- 
tion (2). 

(5)  Make  these  two  expressions  equal  to  one  another 
and  thus  form  your  equation. 

(6)  Solution  of  the  Equation. 

(1)  If  the  equation  contains  fractions,  multiply  each 
term  of  the  equation  by  the  least  common  denominator  of 
the  fractions.     You  now  have  an  equation  with  no  fractions. 

(2)  If  the  equation  contains  parentheses,  do  the  work 
indicated,  collecting  similar  terms  on  each  side  of  the  equation. 

(3)  Now  add  to  or  subtract  from  both  sides  of  the  equa- 
tion, bring  the  unknown  numbers  to  one  side  of  the  equation 
and  the  known  numbers  to  the  other. 

(4)  Divide  both  sides  of  the  equation  by  the  coeflficient 
of  the  unknown  number. 

(5)  The  value  of  the  unknown  number  answers  the 
question  asked,  if  only  one  number  is  called  for.  If  several 
values  are  called  for,  find  the  values  of  these  by  substituting 
in  the  algebraic  expressions  which  you  let  stand  for  these 
numbers  at  the  opening  of  the  statements. 


164  ALGEBRA  —  FIRST  COURSE  txii.  §  78 

(6)  Check  by  substituting  the  value  or  values  of  the 
number  called  for  in  the  original  problem  (not  the  equation 
formed  nor  the  algebraic  expressions  written  down,  for  they 
may  be  wrong)  and  ascertain  by  a  process  entirely  indepen- 
dent of  the  algebraic  solution  whether  every  condition  made 
in  the  English  statements  holds.  This  is  usually  a  pure 
arithmetic  process. 

Problems. 

1.  The  circumference  of  the  front  wheel  of  a  wagon  is  2 
feet  less  than  that  of  the  hind  wheel.  In  going  14  miles  the 
front  wheel  makes  1.4  times  as  many  revolutions  as  the 
hind  wheel.  How  many  feet  are  there  in  the  circumference 
of  each  wheel  ? 

2.  The  rate  a  boatman  can  row  in  still  water  is  1  mile 
more  than  twice  the  rate  of  the  stream,  and  it  takes  3  times 
as  long  to  row  12  miles  upstream  as  it  takes  to  row  10  miles 
downstream.  What  is  the  rate  of  the  rowing  and  the  rate 
of  the  stream  ? 

3.  A  tank  has  two  pipes,  one  through  which  water  flows 
in  and  the  other  through  which  water  flows  out.  The 
volume  flowing  out  per  hour  when  the  emptying  pipe  is  open 
is  1 1  gallons  less  than  the  volume  flowing  in  when  the  filling 
pipe  is  open,  so  that  it  takes  4f  times  as  long  to  empty  the 
tank  as  it  does  to  fill  it.  Find  the  number  of  gallons  per 
hour  which  can  flow  through  each  pipe. 

4.  If  $500  yields  as  much  interest  in  the  same  length  of 
time  as  $750  at  2%  less  rate  of  interest,  what  are  the  rates 
at  which  each  sum  is  drawing  interest? 

5.  In  planting  an  orchard,  a  man  found  that  if  he  planted 
6  more  trees  in  a  row,  he  could  plant  1800  in  the  same  num- 
ber of  rows  that  he  had  originally  planned  for  1500  trees. 
How  many  rows  had  he  planned  for  ? 

6.  A  man  wished  to  sow  a  field  with  wheat,  rye  and  vetch. 
The  price  of  the  wheat  was  10  cents  more  per  hundredweight 


XII.  §78]        EQUATIONS  INVOLVING  FRACTIONS  165 

than  the  price  of  the  rye,  while  the  price  of  the  vetch  was 
4  cents  more  per  pound  than  12  times  the  price  of  the  wheat 
per  pound.  He  arranged  to  pay  for  the  wheat  and  rye  an 
average  price  per  hundredweight.  He  bought  as  many 
hundredweight  of  wheat  and  rye  for  $23.75  as  he  bought 
pounds  of  vetch  for  $4.  What  was  the  price  of  each  kind  of 
grain  ? 

7.  The  number  of  trees  in  a  row  in  an  apple  orchard  is 
10  more  than  3  times  the  number  of  rows.  The  ratio  of 
the  number  of  trees  in  a  row  to  the  number  of  rows  is  5. 
How  many  trees  are  there  in  a  row  ? 

8.  There  is  a  number  the  sum  of  whose  digits  is  11.  If 
the  digits  are  interchanged  and  the  original  number  is  divided 
by  the  new  number  thus  formed,  the  quotient  is  2  with  a 
remainder  of  7.     Find  the  number. 

9.  Two  triangles  whose  areas  are  11 J  square  inches  and  15 
square  inches  respectively  have  equal  altitudes.  The  length 
of  the  base  of  one  is  5  inches  less  than  twice  the  length  of  the 
base  of  the  other.     What  is  the  length  of  the  base  of  each  ? 

10.  What  number  must  be  subtracted  from  the  denomi- 
nator of  the  fraction  ^^  to  increase  the  value  of  the  fraction 
by  V? 

11.  A  boy  can  run  two  and  a  half  times  as  fast  as  he  can 
walk,  and  he  can  run  100  yards  in  15  seconds  less  than  he 
takes  to  walk  it.  Find  his  speed  in  yards  per  second,  when 
running  and  when  walking. 

12.  A  football  player  starts  with  the  ball  from  the  middle 
of  the  field  for  a  touchdown.  An  opposing  player,  who  can 
run  2  yards  a  second  faster,  starting  10  yards  behind  him, 
catches  him  on  the  20-yard  line.    How  fast  did  each  man  run  ? 

13.  When  20  grams  of  a  certain  liquid  are  mixed  with  5 
grams  of  another  liquid  whose  density  is  1.5  greater  than 
that  of  the  first  liquid,  the  density  of  the  mixture  equals  the 
product  of  the  densities  of  the  two  liquids.  What  is  the 
density  of  each  liquid  ? 


CHAPTER  XIII 
'  QUADRATIC   EQUATIONS 

79.  Definitions.  In  Chapter  X  and  in  Chapter  XII  we 
solved  a  number  of  equations  for  the  value  of  an  unknown 
number.  This  number  was  first  represented  by  a  letter, 
then  an  equation  was  formed  from  which  the  numerical 
value  for  which  the  letter  stood  could  be  found.  In  all 
cases  only  the  first  power  of  the  letter  representing  the  un- 
known was  present  in  the  simplified  form  of  the  equation. 
Our  equations  could  all  be  reduced  finally  to  the  simple  form 

ax  -\-  h  =  Of 

where  x  is  the  unknown  number  and  a  and  h  are  known 
numbers. 

We  pass  now  to  equations  in  which  the  square  of  the 
unknown  number  also  occurs;  that  is,  to  equations  of  the 
general  form 

ax^  -\- hx  -{-  c  =  0, 

X  being  the  unknown,  and  a,  h,  c  the  known  numbers.     Either 
6  or  c  may  be  zero,  but  a  must  not  be  zero.     Such  an  equa- 
tion is  called  a  quadratic  equation  in  x;  the  expression 
ax^  +  6x  +  c 

is  called  a  quadratic  expression  in  x. 

Our  problem  now  is,  to  find  the  values  of  x,  which  will 
reduce  this  expression  ax^  +  bx  +  c  to  zero.  This  is  called 
''solving  the  quadratic  equation  ax^  -{-  hx  +  c  =  0  for  x." 

80.  Some  Problems  Leading  to  Quadratic  Equations. 

Example  1.    The  product  of  a  number  by  a  number  2  units  less  is  8. 
What  is  the  number? 
Solution : 

Let  n  =  the  number. 

166 


XUI.I80]  QUADRATIC  EQUATIONS  167 

Then  n  —  2  =  the  number  2  units  less, 

and  n  (n  —  2)  =  the  product  of  the  numbers; 

but  8  =  the  product  of  the  two  numbers. 

n(n-2)=8.     Why? 
We  are  to  find  the  value  of  n  to  verify  this  equation. 
Multiplying  as  indicated 

n2  -  2  n  =  8. 
Taking  8  from  both  sides  of  the  equation,  we  have 

n2  -  2  n  -  8  =  0. 
This  is  a  quadratic  equation  in  n,  from  which  n  is  to  be  found. 
Factoring,  we  have 

(n  -  4)  (n  +  2)  =  0. 
Here  we  have  the  product  of  two  numbers,  namely  n  —  4  and  n  +  2, 
which  product  is  equal  to  zero.     Therefore  one  or  the  other  of  these 
numbers  is  zero.     Why?     Can  the  product  of  two  numbers  be  zero 
without  one  of  the  numbers  being  zero  ? 
Therefore  we  conclude  that 

either    n  —  4  =  0 
or    n  +  2  =  0. 
Let  n  -  4  =  0. 

Then  n  =  4,  one  of  the  numbers, 

n  —  2  =  2,  the  other  number. 
Check:   Since  one  of  the  numbers  is  4  and  the  other  number  is  2, 
the  second  is  2  less  than  the  first,  so  this  statement  is  satisfied  by  the 
numbers  4  and  2.     The  product  of  the  two  numbers  is  4  •  2,  which  is  8. 
So  this  statement  is  satisfied  and  4  and  2  are  correct  answers. 
Let  the  other  factor  equal  zero: 

n  +  2  =  0. 
Then  n  =  —  2,  one  of  the  numbers, 

and  n  —  2  =  —  4,  the  other  number. 

Check:  Since  one  of  the  numbers  is  —  2  and  the  other  is  —  4,  the 
second  number  is  2  less  than  the  first,  and  their  product  is  ( —  2)  ( —  4), 
which  is  8.     So  —  2  and  —  4  are  correct  answers  to  our  problem. 

Example  2.     The  height  of  a  rectangle  is  2  cm.  less  than  its  base. 
Its  area  is  8  square  centimeters.     What  are  its  dimensions? 
Solution : 

Let  b  =  the  number  of  units  in  the  length  of  the  base; 

then  6  —  2  =  the  number  of  units  in  the  length  of  the  altitude, 

so  6  (6  —  2)  =  the  number  of  square  units  in  the  area; 

but  8  =  the  number  of  square  units  in  the  area. 

6  (6  -  2)  =  8.     Why? 


168 


ALGEBRA  —  FIRST  COURSE 


[XIII.  §  80 


This  is  the  same  equation  as  we  had  in  Example  1,  so  we  proceed  as 
before. 

Multiplying  out: 

fe2  -  2  6  =  8. 
62  —  2  6  —  8  =  0,  subtracting  8  from  both  members. 
(b  -  4)  (6  +  2)  =  0,  factoring. 
Therefore,  either  6  —  4  =  0  or  6+2  =  0.     Why? 

If  6-4  =  0, 

6  =  4,  the  number  of  cm.  in  the  length  of 
the  base. 
6  —  2  =  2,  the  number  of  cm.  in  the  length  of 
the  altitude. 
Check:  These  values  satisfy  the  condition  of  the  problem  that  the 
j^  altitude  is  2  cm.  less  than  the 

base   and  also    the    condition 
that  the  area  must  be  8  sq. 
cm.,  since  2  •  4  gives  8,  which 
is  the  area  of  the  rectangle. 
But  suppose  that 
6  +  2=0. 
b       Then  6  =-2,  the  num- 

ber of  cm.  in  the  length  of  the 
base; 

and  6  —  2  =  —  4,  the  num- 

ber of  cm.  in  the  length  of  the 
altitude. 

Check:  These  values  satisfy 
the  condition  of  the  problem 
that  the  altitude  is  2  cm.  less 
than  the  base,  and  also  the  condition  that  the  area  is  8  sq.  cm., 
since  -  2-  (-  4)  =8. 

So  we  have  two  rectangles  that  answer  the  conditions  of  the  prob- 
lem. 

The  graphic  representation  of  this  is  shown  in  the  figure.  The  two 
rectangles  differ  only  in  their  position  with  respect  to  the  origin. 

Exercises. 

1.  A  rectangle  is  11  cm.  longer  than  it  is  wide,  and  con- 
tains an  area  of  26  sq.  cm.  What  are  the  length  and  width 
of  the  rectangle? 

2.  A  rectangle  is  5  cm.  longer  than  it  is  wide  and  contains 


XIII.  §  81]  SQUARE  ROOTS  —  PYTHAGOREAN  THEOREM     169 

—  6  sq.  cm.  of  area.     What  is  the  number  of  cm.  in  its 
width?     In  its  length? 

3.  A  parallelogram  has  an  area  of  60  square  units.  Its 
base  is  11  cm.  longer  than  its  corresponding  altitude.  What 
is  the  length  of  the  base  ?     Of  the  altitude  ? 

4.  A  square  has  an  area  of  625  sq.  cm.  What  is  the  length 
of  its  side?  (Be  sure  that  you  find  two  numbers  that  can 
be  the  length  of  the  side  of  this  square.) 

5.  Find  the  length  and  width  of  a  rectangle  whose  area 
is  49  sq.  in.  and  whose  length  is  7  in.  more  than  6  times  its 
width. 

6.  Find  the  length  of  the  base  and  the  altitude  of  a  tri- 
angle whose  area  is  10  sq.  in.  and  the  length  of  whose  altitude 
is  3  less  than  2  times  its  base. 

7.  If  the  altitude  of  a  parallelogram  be  diminished  by  4 
units  the  base  would  be  2  units  less  than  its  altitude,  and  its 
area  would  be  24  square  units.  What  is  the  length  of  base 
and  altitude  of  parallelogram  ? 

8.  The  altitude  of  a  parallelogram  is  30  in.  less  than  9 
times  its  corresponding  base.  The  area  is  22  sq.  in.  What 
is  the  length  of  the  base  and  altitude  of  the  parallelogram? 

9.  Write  three  problems  about  dimensions  and  areas  which 
will  lead  to  quadratic  equations. 

81.  Square  Roots  of  Numbers.  Until  this  time,  gener- 
ally we  have  made  use  of  numbers  expressing  measurement 
of  quantities  having  a  common  unit  of  measure.  An  ex- 
ception to  this  was  the  ratio  of  the  circumference  to  its 
diameter.  In  this  case  we  introduced  a  new  character  to 
express  the  ratio. 

There  are  other  geometric  figures  which  contain  lines  for 
which  there  can  be  obtained  no  common  measuring  unit. 
Lines  which  may  be  measured  by  a  common  unit  are  called 
commensurable  lines  (the  word  means  common  measure). 
Lines  which  cannot  be  measured  by  a  common  unit  are  called 
incommensurable  lines  (the  word  means  no  common  measure). 


170 


ALGEBRA  —  FIRST  COURSE 


(XIII,  §  81 


XIII,  §  81]  SQUARE  ROOTS  —  PYTHAGOREAN  THEOREM     171 

We  shall  investigate  such  lengths  through  the  following 
theorem. 

Pythagorean  Theorem.*  The  square  on  the  hypotenuse  (the 
side  opposite  the  right  angle)  of  a  right-angled  triangle  is 
equxil  to  the  sum  of  the  squares  on  the  other  two  sides. 

Draw  the  right-angled  triangle  ABC  (p.  170).  Draw  the 
squares  BAFG,  CBHK,  ACDE  on  the  sides  AB,  BC,  CA  re- 
spectively. We  shall,  by  paper  cutting  and  comparison,  show 
that  square  BAFG  =  square  CBHK  +  square  ACDE.  On 
stiff  paper  draw  a  square  equal  to  the  square  ACDE.  Ad- 
joining it,  as  shown  in  Fig.  2,  draw  a  square  equal  to  square 
G 


Fig.  3. 

CBHK.  Cut  out  the  figure  ABHKDE.  On  the  side  AB  lay 
off  a  distance  AR  equal  to  side  BC.  Draw  the  lines  ER  and 
RH.  Cut  the  figure  along  these  lines.  Draw  a  square  equal 
to  the  square  BAFG,  and  cut  it  out.  On  this  square  ar- 
range the  figure  ERHKD  so  that  ER  shall  fall  on  GB,  and 
RH  on  BA,  as  shown  in  Fig.  3.  Now  place  triangle  BHR 
so  that  HR  shall  fall  on  AF.  Place  triangle  ARE  so  that 
RE  shall  fall  on  FG.  Can  you  explain  by  examining  the 
figures  why  these  parts  fit  so  well  together?  This  shows 
the  correctness  of  the  theorem. 

*  This  theorem  takes  its  name  from   the   Greek  mathematician 
Pythagoras  (580-500  b.c),  who  first  proved  the  theorem. 


172  ALGEBRA  — FIRST  COURSE  [Xiii.§8l 

Let  the  letters  a  and  h  denote  the  lengths  of  the  sides  of 
the  triangle  and  let  c  denote  the  length  of  the  hypotenuse. 
Then  our  theorem,  stated  as  an  equation,  is 

Example  1.     If  a  =  3,  6  =  4,  c  =  ? 
Solution:  Substituting  the  values  given, 
c2  =  32  -|_  42 
=  25. 
From  which  c  =  5. 

There  is  a  second  value,  c  =  —  5,  which  we  discard,  because  we  are 
considering  the  sides  of  our  triangle  as  measured  by  positive  numbers. 
Draw  this  triangle  and  check  by  measurement. 

Exercises.     Similarly  solve  and  check  the  following: 

1.  If  a  =  5,     6  =  12,  c  =  ?      3.   If  a  =  16,  c  =  34,  6  =  ? 

2.  If  a  =  15,  6  =  8,     c  =  ?      4.    If  c  =  f ,     6  =  1,    a  =  ? 

Example  2.     If  a  =  3,   6  =  5,  c  =  ? 
Solution:  c^  =  a^  -{-  h^. 

Substituting  a  =  3,  6  =  5, 

C2   =  32  +  52 

=  34. 

Then  c  =  that  number  whose  square  is  34. 

This  number  is  called  the  square  root  of  34. 

Now  there  is  no  number  such  as  we  have  had  up  to  this  time,  by 
which  we  can  express  the  exact  root  of  34.  In  other  words,  the  two 
lines,  namely  the  hypotenuse  and  either  side  of  this  triangle  are  incom- 
mensurable. It  is  necessary  to  introduce  a  new  symbol  to  express 
this  new  number.    We  write 

c  =  34^  or  V34. 

These  two  symbols  mean  the  same  thing.  We  may  use  them  inter- 
changeably. The  first  is  read  "34,  exponent  i,"  and  the  second  is 
read,  "the  square  root  of  34."  There  would  be  no  objection  to  reading 
the  first  as  "the  square  root  of  34." 

Since  c  •  c  =  34,  we  must  have 

34* .  34*  =  34; 
also  V34  •  V34  =  34. 

Such  numbers  are  called  incommensurable  or  irrational  numbers  to 
distinguish  them  from  the  numbers  which  we  have  been  using,  which 


XIII.  §  82]  SQUARE  ROOTS  —  PYTHAGOREAN  THEOREM     173 

are  called  rational  numbers.  The  expression  of  the  ratio  of  commen- 
surable lines  is  a  rational  number.  The  expression  of  the  ratio  of 
incommensurable  lines  is  an  irrational  number. 

82.  Geometric  Construction  of  the  Square  Root  of  a 
Number. 

Problem.  To  draw  a  square  the  length  of  whose  side  is 
expressed  hy  an  irrational  number. 

Although  we  cannot  always  express  the  length  of  the  side 
of  a  square  in  terms  of  a  desired  unit,  we  can  always  draw 
the  square  after  finding  the  length  of  its  side  geometrically. 

Example  1.     Draw  a  square  whose  area  is  17  square  units. 

To  do  this  we  decide  upon  two  numbers  which  are  squares  whose 
sum  is  17.  These  numbers  are  16  and  1.  So  count  4  to  the  right  and 
1  up.     Thus: 


■1 

^ 

""^ 

o\ 

[—^ 

If 

\/Tr 

0 

•17 

The  line  joining  the  origin  to  the  point  last  counted  is  the  side  of 
the  square  called  for  because  the  square  on  it  is  equal  to  the  sum  of  the 
squares  on  the  other  two  sides  of  the  right-angled  triangle  one  of  which 
is  4  and  the  other  1. 

Example  2.     Draw  a  square  whose  area  is  21  square  units. 

In  this  case  the  nearest  that  we  can  come  to  finding  two  squares 
whose  sum  is  21  is  16  and  4,  which  added  give  us  20.  It  will  then  be 
necessary  first  to  find  the  length  of  the  side  of  a  square  whose  area  is 
20,  and  then,  using  this  for  one  side  of  our  right  triangle  and  1  as  the 
length  of  the  other  side,  we  shall  get  the  length  of  the  side  of  the  square 
called  for.  In  taking  the  length  of  20'  from  one  figure  to  use  in  the 
other  it  is  better  to  use  our  compass  as  we  cannot  measure  it  by 
means  of  the  units  on  our  measuring  line.    Why? 


174 


ALGEBRA  — FIRST  COURSE 


[XIII.  §  82 


v'ij 

21 

0 

/r 

Example  3.     Draw  the  square  whose  area  is  28  square  units. 
Examining  this  number  you  will  find  that  it  may  be  separated  into 
two  factors,  one  of  which  is  a  perfect  square  while  the  other  is  not. 

28  =  4  •  7. 
Then  28^  =  4' .  7^  =  2  .  7^     (Why?) 

That  is,  we  have  the  law 

(a6)'  =  ah^,  or,  \^ah  =  Va-  Vb, 

Test  to  see  if  this  is  true  by  letting  a  =  9,  6  =  4. 

Does  9*-4^  =  (9-4)*? 

So  in  our  drawing  for  the  side  of  the  square  whose  area  is  28,  we  find 
the  length  of  the  side  of  a  square  whose  area  is  7,  and  take  twice  this 
length  and  draw  our  square  which  wiU  be  equal  to  4  times  the  square 
of  7*. 


f 

3 

ri 

b 

\ 

n   0 


2^11 


\f7-- 


re  0 


28 


TT 


2/7 


Exercises.     Draw  the  squares  whose  areas  are: 
13;        27;        37;        48;        23. 


XIII.  §  82]  SQUARE  ROOTS  —  PYTHAGOREAN  THEOREM     175 


The  square  roots  of  numbers  will  arise  often  in  the  solu- 
tion of  quadratic  equations.  Until  this  time  we  have  care- 
fully avoided  such  equations. 

Example  4.  Suppose  you  were  asked  to  solve  the  very  simple 
quadratic  equation 

x2  -  2  =  0. 

According  to  the  methods  previously  used,  you  would  factor  it  into 
the  sum  and  difference  of  the  roots  of  x^  and  2.  Try  it.  You  are 
asked  to  find  the  square  root  of  2. 

The  geometric  solution  is  very  easily  obtained.     You  have  but  to 
count  1  unit  to  the  right  and  1  unit  up  and  join 
this  point  to  the  origin,  or,  what  amounts  to 
the  same  thing,  the  diagonal  of  a  square  whose 
side  is  1  unit  is  2^. 

Draw  a  square  whose  side  is  1  decimeter 
in  length.  Draw  its  diagonal.  With  your  com- 
pass lay  off  on  the  diagonal  a  distance  equal  to 
the  side  of  the  square.  How  often  can  you 
lay  it  off? 

Take  one-tenth  of  the  side  of  the  square  and 
lay  it  off  as  many  times  as  you  can.     How  often  can  you  lay  this  off? 

Take  one  hundredth  of  the  side  of  your  square  and  lay  it  off.  How 
often  can  you  lay  it  off? 

In  the  first  trial  you  get  1.  In  the  second  trial,  14.  In  the  third 
trial,  141.  If  you  would  use  one-thousandth  part  of  the  side  of  the 
square,  you  would  get  1414.  That  is,  you  find  that  as  you  diminish  your 
measuring  line  you  come  closer  to  the  square  root  of  2. 


/ 

p  =  l, 

1.42  =  1.96, 
1.412  =  1.988  -h, 
1.4142  =  1.999  -f- 
and  so  on. 

Since  it  is  true  that  (-  1)^  =  1;  (-  1.4)2  =  1.96;  (_  1.41)2  = 
1.988;  then  —  1,  —  1.4,  —  1.41,  —  1.414  and  so  on  are  numbers 
approaching  closer  and  closer  to  a  number  whose  square  is  2.  So  there 
are  two  numbers  whose  square  is  2. 

But  we  can  never  arrive  at  the  exact  value  of  these  nimibers  either 
in  the  form  of  a  fractional  number  or  as  a  terminating  decimal.  We 
must  content  ourselves  therefore  by  merely  indicating  them  by  the 


176  ALGEBRA  — FIRST  COURSE  txill,  §  84 

symbol  2*  or  by  V2  for  the  positive  square  root  and  by  —  2^  or  —  V2 
for  the  negative  square  root. 

Similarly,  the  positive  number  whose  square  is  a  is  indicated  by  Va 
or  a*.  The  negative  number  whose  square  is  a  is  indicated  by  the 
symbol  —  Va  or  —  (a)^  The  symbol  V  is  called  the  radical  sign. 
Either  one  of  the  two  numbers  whose  square  is  a,  that  is  Va  or  —  Va, 
is  called  the  square  root  of  a;  they  are  distinguished  as  the  positive 
square  root  and  the  negative  square  root  respectively. 

We  can  now  solve  our  equation 

a;2  -  2  =  0. 

Factoring:       {x  -  2^)  {x  +  2^)  =  0. 

Hence  x  =  2^  or  V2, 

and  x  =  -  2^  or  -  V2. 

These  are  usually  written  together: 

X  =  ±  V2. 
Read  this  "a:  equals  the  positive  or  the  negative  square  root  of  2." 

83.  Rational  and  Irrational  Numbers. 

Definitions.  Numbers  such  as  Vs,  Vs,  V?,  and  so 
on,  whose  values  can  be  expressed  approximately,  but  not 
exactly,  by  a  numerical  fraction  or  a  terminating  decimal, 
are  called  irrational  numbers. 

So  far  the  student  has  dealt  almost  entirely  with  numbers 
which  were  either  integral,  positive  or  negative,  or  quotients 
of  such  integers,  that  is,  numerical  fractions.  All  such  num- 
bers are  called  rational  numbers. 

84.  Factoring  of  Quadratic  Expressions.  Consider  the 
quadratic  expression 

2  a2  4-  7  a  +  4. 

No  matter  how  long  we  try,  we  cannot  find,  by  means  of 
the  processes  already  given,  the  factors  of  this  expression. 
The  reason  for  this  is  that  the  factors  contain  irrational 
numbers,  and  the  factoring  which  we  have  previously  con- 
sidered involves  only  rational  numbers.  We  shall  now 
investigate  the  drawings  by  which  we  may  manage  to  arrive 


XIII.  §  84]     FACTORING  OF  QUADRATIC  EXPRESSIONS      177 

at  the  factoring  of  this  expression.  However,  we  shall  not 
start  our  investigation  with  this  more  complicated  form, 
but  we  shall  start  with  the  expression  a^  -j-  5  ^^  -f-  6  (p.  130 
Ex.  6),  and  show  that  by  a  longer  process  we  may  draw  this 
as  a  rectangle  and  find  its  sides  to  be  the  same  as  we  arrived 
at  by  the  former  plan. 

While  we  factor  this  simpler  expression  by  this  method 
for  the  sake  of  practice,  we  do  not  use  it  in  actual  work  when 
the  factors  can  be  given  readily  by  the  shorter  plan. 

Example  1.     To  factor        a^  -\-  5  a  -\-  Q; 

that  is,  to  find  the  dimensions  of  the  rectangle  of  which  it  expresses 
the  area. 

Note.  See  that  you  follow  the  directions  and  answer  the  questions 
asked  —  each  one  finished  before  reading  the  next. 

The  first  term  of  the  expression  is  a  2.  What  are  the  dimensions  of 
the  square  which  represents  this  term?     Draw  it.     The  next  term  ia 


a+'y 

?-> 

^ 

a-h'4 

%a 

'V4 

+ 

a 

+ 

a' 

'A  a 

V2 

0] 

9 

i 

c 

I 

a+ 

'A 

V 


a+'A-V, 


5  a.     Now,  instead  of  figuring  out,   as  formerly,  two  different  sized 

rectangles  whose  sum  is  5  a,  add  to  the  two  sides  of  the  square  two 

rectangles  of  the  same  size,  whose  sum  is  5  a.     Of  course,  the  area  of 

5a  5  . 

each  is  -^ ,  the  dimensions  being  a  and  -  •    Having  drawn  this,  exam- 

ine  your  figure,  and  determine  what  area  it  will  be  necessary  to  add 


178  ALGEBRA  — FIRST  COURSE  [xiii,  §  84 

in  order  to  make  a  complete  square.  What  are  the  dimensions  of  the 
area  added  ?  What  are  the  dimensions  of  your  figure  after  adding  this 
area?  This  process  is  spoken  of  as  completing  the  square.  Is  the  area 
of  the  square  you  now  have  the  same  as  that  given  by  the  original 
expression,  a^  +  5a  +  6?  How  much  more  area  has  been  added? 
What  is  it  necessary  to  do  to  bring  back  the  original  area? 

Since,  in  order  to  complete  the  square  we  used  an  area  of  Qj  square 
units,  and  our  expressiong  ave  us  an  area  of  only  6  square  units  it 
will  be  necessary  to  subtract  |  square  unit.  Subtract  a  square  whose 
area  is  |.  We  now  have  the  difference  of  two  squares  as  in  the  type 
form  a^  —  h^.  See  p.  134.  Proceed  as  in  the  drawing  for  that  type 
form.  You  produce  a  rectangle,  one  side  of  which  is  a  +  f  +  ^  and 
the  other  is  o  +  |  —  ^,  or,  collecting  terms,  one  side  is  a  +  3  and  the 
other  is  a  +  2. 

The  algebraic  expression  for  this  is: 

a2  +  5a  +  6  =  a^  +  5a -\--\^- -  i 
=  (a  +  3)  (a  +  2). 

Examine  the  figure  which  you  drew  for  the  same  expression  in  (Ex.6, 
p.  130),  and  satisfy  yourself  that  the  resulting  rectangles  are  identical. 

Exercises.  As  in  the  above  illustration,  draw  and  write  the 
algebraic  expression  for  the  factors  of  the  following.  State  the 
values  of  the  leading  letter  that  will  make  the  expression  zero. 
Check  the  correctness  of  these  values  by  substituting  each 
in  the  original  expression  and  doing  the  work  indicated  to 
see  that  it  gives  zero. 

1.  m2+llm  +  28.  3.   s^+iOs  +  lG. 

2.  r2  +  7  r  +  6.  4..   a^ -{- 9  ah  +  U  h\ 

Example  2.     We  are  now  ready  to  find  the  dimensions  of  the  rec- 
tangle whose  area  is  given  by  the  expression 
2a2-j-7a  +  4. 

Since  the  first  term  is  not  a  perfect  square,  we  make  it  a  square  by 
multiplying  the  entire  expression  by  2  (the  least  number  that  will  make 
it  a  square).  The  rectangle  that  we  draw  will  have  an  area  of  twice 
the  area  of  the  original  rectangle.  Wc  shall  rectify  this  by  taking  a 
rectangle  which  has  one  of  its  dimensions  one-half  of  the  one  we  draw. 
2  a2  +  7  a  +  4  =  H4  a2  +  14  a  +  8). 

Proceed  as  in  the  preceding  example;  draw  a  square  whose  area  is 
4  a';   add  two  rectangles  the  sum  of  whose  areas  is  14  a,  each  having 


XIII.  §  84]     FACTORING  OF  QUADRATIC  EXPRESSIONS      179 

one  dimension  2  a  and  the  other  |;  complete  the  square  by  adding  a 
square  whose  area  is  ^-.  In  doing  this  you  have  used  V"  naore  than 
was  given  in  the  expression  4  q^  -|-  14  a  +  8,  so  you  must  subtract  a 
square  whose  area  is  -V-.     That  is,  you  must  subtract  a  square  whose 

side  is  —rr-,  which  may  be  written  5  Vl7. 

The  drawing  is  as  follows: 


2a+| 


2a 


2 


2a+ 




-\ 

7a 

AS. 

u 

7a 

4( 

1' 

■^H 

l 

1 

tf 

1 

2 

I 

1 

_i 

, 

2a 

i 

U 

/If 

"JiW 


The  algebraic  expressions  are: 

2  a2  +  7  a  +  4  =  M4  a^  +  14  o  +  8) 

=  |(4a2  +  i4a+-^--i^)  __ 

=  l(2a  +  I  -  I  Vl7)  (2a  +  I  +  ^  Vl7). 

If  we  wish  to  know  what  values  of  a  will  make  2  a^  +  7  a  +  4  zero, 
we  make  each  factor  in  turn  equal  to  zero  and  find  the  value  of  a. 
Letting  2  a  +  ^  -  ^  VT7  =  0, 

2a=-l  +  |  VT7» 
a  =  -  1  +  i  Vl7. 
Substituting  this  value  into  the  given  expression  to  see  if  it  gives 

zero,  we  have       

2  (-1  +  i  Vl7)2  +  7  (  -  I  +_i  Vl7)  +4  _ 

=  2  at  -  I  V^17  +  il)  +  (-^  + 1  Vl7)  +  4 
=  (¥•-!  Vl7  +-V)  +  C-¥  +  i  vl7)+  4 
«0. 


180 


ALGEBRA  — FIRST  COURSE 


[XIII,  §  84 


Since  we  were  solving  for  a  number  that  would  make  the  expression 
zero,  —  I  +  J  Vl7  must  be  a  correct  answer. 

By  making  the  other  factor  equal  to  zero  let  the  student  find  an- 
other value  for  a  that  wUl  make  the  given  expression  zero,  and  check. 

Exercises.  By  the  process  just  given,  factor  and  find  the 
value  of  the  letter  that  will  make  the  expression  zero. 

1.  3s2  +  7s  +  2.  3.  m2  +  3m  +  l. 

2.  8a;2  +  10a;+l.  4.   5r2  +  9r  +  3. 
Example  3.     Factor  the  expression  2a^  -{-Sa  —  5. 

We  write  2a2  +  3a-5  =  H4a2  +  6a-  10). 

Directions:  Draw  the  square  whose  area  is  4  a^.  Divide  the  area 
6  a  into  two  equal  rectangles,  one  dimension  of  each  being  2  a.  Add 
the  rectangles  to  the  square  as  in  the  preceding  example.  Determine 
the  area  necessary  to  complete  the  square  and  add  to  the  figure. 

Notice  these  facts:  In  the  first  place  you  did  not  have  an  area 
equal  to  4  a^  +  6  a  and  yet  you  have  used  that  much  in  your  drawing. 


sa+h- 

-Vz 

\^ 

\ 

1 
r 

1 

1 

^72  1 

[s^ 

9/1 4-- 

14- 

1 

'& 

1   ■ 
1 

1 

1 

1  _ 

1 

1 

1 

h9 

1 

1 

1  " 

1 
1 

k 

- 1 
1 

1 

I 

4. 

■■^ 

1 

0 

2i 

1     ia-^% 

You  lack  10  square  imits  of  having  that  area,  and  in  addition  to  this, 
in  order  to  complete  the  square,  you  used  \  square  units.  All  together 
you  used  -V-  square  units  more  than  was  given  you,  so  that  you  must 
subtract  -V-  square  units.  Subtract  this  in  the  form  of  a  square  and 
proceed  as  in  the  above  examples. 

By  taking  a  rectangle  whose  base  is  the  same  as  this  one  and  whose 
altitude  is  ^  of  the  altitude  of  this  one,  we  shall  have  the  rectangle 
called  for. 


XIII.  §  84]     FACTORING  OF  QUADRATIC  EXPRESSIONS      181 


The  algebraic  expression  for  this  is 


2a2  +  3a  -5  =  ^ia'^  +  Qa  -10) 
=  H4a2  +  6a  +  f - 
=  H2a  +  5)(2a 
=  (2a +  5)  (a  -1). 
Example  4.     Factor  the  expression  6  a^  —  a  —  12. 


-¥) 


2) 


Multiplying  the  expression  by  6  in  order  to  make  the  first  term  a 
square: 

6a2-a-12  =  i  (36a2  _  6a  -  72). 
As  in  the  preceding 


6a-H-'y^ 


example,  draw  a  square 
whose  area  is  36  a^ 
square  units.  Subtract 
two  equal  rectangles 
the  sum  of  whose  areas 
is  6  a.  See  p.  132. 
Why  subtract  instead 
of  add? 

After  subtracting 
one  of  these  rectangles 
from  your  square,  you 
will  find  that  it  is  nec- 
essary to  add  a  square 
of  area  I  before  you 
can  subtract  the  second 
rectangle,  so  that  when 
you  have  subtracted 
the  two  rectangles,  you 
have  a  square  which 
contains  |  more  than 
an  area  represented  by 
36a2-6a.  This  area 
is  72  square  units  more 
than  the  area  repre- 
sented by  the  expres- 
sion 36  a2  -  6  a  -  72. 
So  that,  in  order  to 
have  an  area  equal  to 
that  called  for  by  the 

expression  36  a^  —  6  a  —  72,  it  will  be  necessary  to  subtract  a  square 
whose  area  is  equal  to  the  sum  of  72  and  I,  that  is,  an  area  of  ^f  ^  square 
units.     Subtract  this  square  and  then  proceed  as  before. 


•\ H 


H h 


6a 


•i — I — I 1 — I- 


182  ALGEBRA  —  FIRST  COURSE  [xiii,  §  85 

The  algebraic  expression  for  this  is 

6a2  _  a  -  12  =  I  {SQa^  -  6a  -  72) 

=  H36a2-6a  +  i-^|^) 
=  i(6a-9)(6a  +  8) 
=  (2  a  -3)  (3  a +  4). 

Exercises.  According  to  the  preceding  illustrations  draw 
and  write  factors  for  the  following;  find  values  of  leading 
letters  that  will  make  the  expression  zero;  check. 

1.  2r2-3r-14.  6.  4:X^-\-7x-S, 

2.  3h^-10h-d.  6.  Sx^-4.x-l. 

3.  lOr^  -  27 rs  + 5  s\  7.  S  x"" -\- 4: x  -  1. 

4.  6r2  +  17r-3.  S.  Sx^-5x+l. 

85.  Factors  of  «a?^  -]r  bx  -\-  c.  We  now  consider  the  fac- 
toring of  the  general  quadratic  expression 

ax^  -{-hx  -\-  c, 

and  the  finding  of  the  values  of  x  that  will  make  the  expres- 
sion zero.  Since  this  is  the  general  form  for  all  the  expres- 
sions given  above,  the  treatment  is  the  same. 

Make  the  first  term  a  square  by  multiplying  the  expression 
by  a: 

ax^  +  6x  +  c  =  -  (a^x^  +  ahx  +  ac). 

The  area  drawn  is  a  times  the  correct  value;   we  take  one 

ath  part  of  it. 

Draw  a  square  whose  area  is  a^x^.     Its  side  is  ax.     Draw 

dbx 
two  rectangles  the  simi  of  whose  areas  is  ahx.     Each  is  -^  • 

Their  dimensions  are  ax  and  ^.     Complete  the  square  by 

adding  a  square  whose  area  is  -j.     You  had  the  area  ac,  you 

52 
used  area  -j ,  therefore  you  must  subtract  a  square  which  is 


XIII.  §  85]     FACTORING  OF  QUADRATIC  EXPRESSIONS      183 


the  difference  between  these  two  areas,  that  is,  a  square 

7^2 4:  ac 

whose    area    is  j .      The    side    of    this    square    is 

\//^2 4  ac 

^r .     Proceeding  with  the  rest  of  the  drawing  as  in 

the  preceding  cases  of  factoring  the  difference  of  two  squares, 
we  have  the  following  figure : 


ax¥' 


y^^ 


aa7+-|-H g 


ah  X 


a'x' 


ah  X 


0  v/p-^ac 


ax 


> 


ax  + 


The  algebraic  expressions  for  the  various  steps  are: 
ax^  -{-hx  +  c 

=  -  {aV  +  abx  +  ac) 


1/       ,  b      V62-4ac\/       ,  6   ,  V62-4oc\ 

=ar+2 — 2 — Jr+2+-^ — j 


Therefore 


aic^+6ic+c=a(a?  + 


6      Vb^  —  4:ac\f       ,  0   .  VD^  — 4ac^ 
^"^  +  2 2 A^^  +  2  + 

)r+ 2^ i' 


b  +  Vft^  -  4ac\/     .  6  -  Vfe^  _  4 


2a 


184  ALGEBRA  — FIRST  COURSE  [xiii.  §  85 

To  find  the  value  of  x  that  will  make  the  expression  zero, 
we  write  each  factor  in  turn  equal  to  zero  and  solve.     Thus 


let 

&      V62-4ac 

aa;  1  2              ^          ~ "' 

then 

b  ,  V62-40C 
ax-     ^+         2         ; 

-6  + V62-4ac 
2a 

Let 

^^6^V6--4ac^„_ 

then 

—  6  -  V62  -  4  ac 
a;  = , 

2a 

The  student  should  convince  himself  that  these  values  will 
make  the  expression  ax^  -\-  hx  -\-  c  zero,  by  substituting 
them  in  the  expression  and  doing  the  work  indicated.  This 
will  be  sufficient  to  show  that  these  values  are  the  correct 
ones,  even  though  there  may  be  some  limitations  to  our 
drawings. 

These  values  are  generally  written  together  in  the  form 

-  6rfc  V62  -  4 ac 

X  =  jz 

2a 

Since  a,  h,  c  may  take  on  any  values  with  the  limitation  that 
a  cannot  be  zero,  we  may  use  this  answer  as  a  formula  by 
which  we  may  solve  any  quadratic  equation. 
Eor  example,  take  the  expression 

2x^  +  7x  +  4:. 

To  find  the  value  of  x  that  will  make  the  expression  zero,  or, 
in  other  words,  solve  the  equation 

2x^  +  7x  +  4:  =  0. 


XIII.   §86]     SOLUTION  OF  QUADRATIC  EQUATIONS  185 

In  this  equation  a  =  2,  6  =  7,  c  =  4.  Substituting  these 
values  in  the  formula,  we  have 

-7  ±V72- 4.2.4 
^= 2T2 

^  -7±VT7 
4 

This  you  will  recognize  as  the  same  exercise  used  in 
Example  2.  _Compare  the^wo  processes  and  convince  your- 
self that  they  are  the  same.  The  check  is  the  same  as  given 
in  that  exercise. 

86.  Algebraic  Solution  of  the  General  Quadratic  Equa- 
tion. As  has  been  remarked,  there  are  always  some  limita- 
tions to  be  placed  upon  drawings.  The  following  solution  is 
purely  algebraic. 

Consider  the  equation 

ax^  -\-  bx  +  c  =  0, 

where  a,  h,  c  are  given  numbers.     To  find  x  in  terms  of  a,  h,  c. 
First  step.     Subtract  c  from  each  member  of  the  equation. 

ax^  -\-hx  =—  c. 

Second  step.     Divide  by  a, 

x^  -\-  -X  = 

a  a 

Third  step.  Complete  the  square  by  adding  to  each  mem- 
ber the  square  of  half  the  coefficient  of  x,  that  is 

62 


(2  J 


4a2 


2,6      ,h^  -_^_£ 

This  may  be  written 

62  -  4  ac 


hM' 


ia-     "y' 


186  ALGEBRA  —  FIRST  COURSE  [Xiii,  §  86 


Hence  x  +  ^^=+^-^^     or     -V^^^' 

Therefore  the  two  values  of  x  are 

b    ,  Jb^-4:ac 
2  a      V       4  a^ 

b        ,  /62  -  4  ac 
These  values  of  x  may  be  simplified  a  little.     For  we  have 


v/ 


52  _  4^,      V6^-4ac^     ^^^^^ 


4a2  2  a 

Then  the  values  of  x  are 


b    .  Vb^  -  4~ 


+ 


ac 


or         a;  = 


2a  '  2a 

b        V62  -  4  ac 


-  6  +  V62  - 

-  4ac 

2a 

-  6  -  \/62  ■ 

—  4ac 

2a  2a  2a 

More  compactly, 

-  &  it  Vfr^  -  4  ac 
2  a 

Here  we  use  the  positive  sign  before  the  radical  for  one 
value  of  X,  the  negative  sign  for  the  other. 

This  is  the  same  formula  that  we  developed  by  our  draw- 
ing. The  student  should  fix  this  well  in  his  mind.  When 
the  expression  cannot  be  factored  readily  by  inspection,  the 
shortest  way  to  find  the  values  of  the  unknown  quantity  in  a 
quadratic  equation  is  by  means  of  this  formula. 

Exercises. 

Solve  several  of  the  previous  exercises  by  means  of 
this  formula  and  see  that  you  arrive  at  the  same  result  as 
before. 


XIII,  §87]       SOLUTION  OF  QUADRATIC  EQUATIONS         187 

Solve  by  formula  and  check: 

1.  9r2-4r-6  =  0.  m-2    2m+l^llm+l 

2.  3s2- 7s  +  2  =  0.  m+2      m-2  ~  4-mV* 

3.  12m2+llw  +  2  =  0.  4r  +  7     5  -r  ^4r 

4.  50x2 -4a;- 6  =  0.  19     "^3  +  r  ~  9  ' 

5.  2m2  +  26m  -  271  =0.  9.  x^-2ax  =  b^-  a\ 

6.  2  r2  -  267  r  +  3240  =  0.  10.  ar^  +  2ar  =-r^  +  1. 

11.  m^^ —2  m^p-^2  mn—2  m^p  -\-  2  mnp^  —  n^p^  — 171"^  —  n^. 

12.  a2a;2  +  a2  -  hH^  =  2aH-\-b^-^2  b^x. 

13.  6r2  -  2  6r  =  a  -  &. 

87.  Imaginary  Roots. 

Example.     Given  the  equation 

2x2  + 2a; +  3  =  0. 

To  find  the  value  of  x  that  will  satisfy  it. 

In  this  equation  the  special  values  of  a,  b,  c  are 

a  =  2,  6  =  2,  c  =  3. 
Using  our  formula,  one  of  the  values  of  x  is 


-  6  +  V62  -  4  ac 

2a 

-  2  +  V22  -  4  .  2  . 

3 

4 
-2  +  V-20 

4 

-2  +  2- V-5 

4 

-  1  +\/-5 

Examining  this  root  we  find  that  a  new  number  has  arisen  which 
we  have  not  met  before  in  our  work.  We  are  called  >ipon  to  extract  the 
square  root  of  a  negative  number.  But  there  exists  no  real  number 
which,  when  used  as  a  factor  twice,  wiU  give  a  negative  number.  Such 
an  indicated  even  root  of  a  negative  number  is  called  an  imaginary 
number. 

The  symbol  V—  5  we  shall  regard  as  equivalent  to  V5  •  V—  1; 
similarly  with  regard  to  the  square  root  of  any  other  negative  number. 


188  ALGEBRA  —  FIRST  COURSE  [Xiii.§87 

We  can  take  the  square  root  of  5,  either  approximately  or  by  drawing. 
The  symbol  of  interest  to  us  is  V—  1. 

There  is  no  other  way  of  expressing  this  by  means  of  the  symbols 
that  we  have  been  using,  so  mathematicians  have  agreed  to  write  it 
and  read  it  i.    Our  expression  is,  then,  i  V5. 

Definition.  Since  the  square  root  of  a  number  when  used 
as  a  factor  twice  must  give  the  number,  therefore  we  define  i 
as  that  number  whose  square  is  —  1. 

That  is:  i.i=_l, 

i  '  i  •  i  =  —  i. 
i  '  i  '  i  '  i  =  1. 

We  now  have  i  V5  •  i  Vs  =  5  (—  1)  =  —  5,  so  that  i  V5 
is  the  square  root  of  —  5.  Later  we  shall  discuss  this  new 
number  more  fully. 

We  are  ready  now  to  check  our  exercise.     Substituting 

^ in  place  of  x  in  the  first  member  of  the  given 

equation,  we  have 

2(-l  +  iV5)^  .  2(-14^^V5)   ,  ^ 
4  +  2  '^^ 

^1 -2'iV5-|-5(- 1)       -2  +  2«^\/5 

2  "^2 

=  0, 

as  the  equation  states  that  it  should. 
Finish  this  example  by  substituting 

a  =  2,     6  =  2,     c  =  3, 

in  the  second  value  of  x,  that  is,  in 


+  3 


-  6  -  V62  -  4  ac 
X  =  — 


2a 
Check  as  illustrated  above. 


XIII,  §88]         CALCULATION  OF  SQUARE  ROOTS  189 

Exercise.     Find  the  values  of  x  that  will  satisfy  the  fol- 
lowing equations : 

1.  5a:2  +  3x+*l  =0.  4.   Tx^  -  2a;  +  1  =  0. 

2.  a;2  +  X  +  1  =  0.  6.  4a;2  +  X  +  2  =  0. 

3.  Sx^-7x-\-5  =  0.  6.   15a;2  +  7  =  0. 

88.  To  Calculate  the  Square  Root  of  a  Number. 

Example  1.     What  is  the  square  root  of  55,696? 
By  inspection  we  find  that  it  is  more  than  200  and  less  than  300. 
Call  it  200  +  b.    Then  we  are  to  find  b,  so  that 
First  step:  (200  +  by  =  55,696, 

or  2002  +  2  •  200  6  +  62  =  55,696. 

Then  2  •  200  6  +  6^  =  15,696. 

A  trial  value  for  b  is  now  obtained  by  neglecting  b^  and  taking 
2-200  6  =  15,696; 

h       ^5,696      „_   , 

(The  neglecting  of  6^  may  lead  to  too  large  a  value  for  6;  if  so,  this 
will  be  found  out  at  the  next  step.) 

Our  approximate  square  root  is  now  230  +. 

So  we  start  again. 

Second  step:  Find  6  so  that 

(230  +  6)2  =  55,696, 
or  52,900  +  2  •  230  6  +  62  =  55,696. 

2  •  230  6  +  62  =     2,796. 
Again  neglecting  62,  we  find  the  next  trial  value  from 
2 .  230  6  =  2796, 

h      2796      _   . 

Hence  our  square  root  is  236  +. 
Third  step:  (236  +  6)2  =  55,696. 

We  now  find  that  2362  =  55,696; 

hence  6  =  0;  that  is,  236  is  the  exact  square  root. 

The  numerical  work  here  is  usually  written  as  follows: 
First  step:  5,56,96  [236  square  root 

4 

156 

129 


Second  step:  43 

Third  step:  466 


2796 

2796 

0 


190  ALGEBRA  —  FIRST  COURSE  IXIII.§88 

Example  2.  When  the  given  number  is  not  a  perfect  square,  we 
get  by  this  process  as  close  an  approximation  to  its  square  root  as  we 
like. 

Find  to  three  decimal  places  the  square  root  of  29. 


(1) 

29.00,00,00  1 5.385  square  root. 

25 

(2) 

10.3 

4.00 

10.69 

3.09 

(3) 

.9100 

.8552 

(4) 

10.765 

.054800 
.053825 

.000975 

Example  3.     Find  the  square  root  of  131.12,  to  two  decimal  places. 

The  required  square  root  Hes  between  11  and  12.  Hence  we  may 
neglect  the  decimal  point  for  the  present  and  proceed  as  in  the  pre- 
ceding examples. 

Exercises. 

Find  the  square  roots  of  the  following: 

1.  (a)  2304;  (b)  1369;  (c)  8649;  (d)  5776;  (e)  331,776; 
(/)  50,625;   (g)  5,764,801;    (h)  43,046,721;    (i)  123,454,321. 

2.  Calculate  to  two  decimal  places  the  value  of  each  of 
the  following: 

(a)  13^;  (h)  21^;  (c)  VH;  (d)  VEtTiE;  (e)  (2.7)^; 
(/)  V:012r;   (g)  VmQ;   (h)  (.075)^. 

3.  Calculate  each  of  the  following  to  three  decimals: 
(a)  \/l- 

I  Notice  that  y  «  =  — p  =  -q-  ;  so  calculate  Vs  and  divide  by  3.  J 
(5)y|.     (Notice  tut  V^  =  ^  =  ^«.) 
(C)A.     (Notice  that  A  =  2v^.) 


XIII.  §  891  SUMMARY  —  PROBLEMS                           191 

(d)  -^-  (e)  -^^' 

VlO  2  Vl5 

V2+1  V                     V'2+I     V2+1  V2-1      2-1                  y 
/  ^         3 


W 


Vy-  2 

(Notice  that^J-  =  -A_  .  ^  =  3V7+^^  ^^^    N 
\  V7-2\/7-2\/7  +  2         7-4  / 

Vl  +  Vs 

/xT  .•      .u  .       ^2                 V2          V5-V3      Vl0-V6\ 
Notice  that— ^^ —  =  — = pi  •  —^ ;=  = ^ . 

V  V5  +  V3     VE+Vs    V5-V3  2        / 

a/7  +  V5  Vt  -  V2 

89.   Summary. 

An  expression  of  the  form  of  ax^  -\-  hx  -{■  c  is  called  a 
quadratic  expression  in  x. 

An  equation  of  the  form  ax^  +  hx  +  c  =  0  is  called  a 
quadratic  equation  in  x. 

Pythagorean  Theorem.     If  a  and  6  are  the  sides  of  a  right- 
angled  triangle,  and  c  is  the  hypotenuse,  then 
a2  +  62  =  c2. 

This  theorem  is  used  in  the  construction  of  square  roots  of 
numbers. 

Arithmetic  integers  and  fractions,  positive  or  negative, 
form  the  system  of  rational  numbers.  All  other  numbers 
such  as  IT  or  V2  are  called  irrational  numbers. 

Any  rational  number  can  be  expressed  as  the  quotient  of 
two  integers;  irrational  numbers  cannot  be  so  expressed. 


192  ALGEBRA  —  FIRST  COURSE  [Xlll.§89 

Factors  of  ax^  -\-hx  -\-  c\ 
ax^+hx^-c  =  a  \x J[x ^^ 1 

This  formula  is  to  be  used  when  factoring  by  inspection 
cannot  be  done  readily. 

Solution  of  the  equation  ax^  +  6x  +  c  =  0.  The  two 
solutions  are 

-6  +  Vb^-  4ac        ,           -  h  -Vb^-  ^ac 
X  = pr and  X  = ^ 

The  imaginary  unit  i  or  V  —  1  is  defined  as  that  number 
whose  square  is  —  1. 

When  6^  —  4  ac  is  negative,  the  values  of  x  contain  this 
imaginary  unit. 

Problems  Involving  Quadratic  Equations.* 

In  the  following  problems  translate  the  English  into  alge- 
braic expression  according  to  instructions  given  on  p.  162. 
Form  the  equation  and  if  it  is  quadratic  add  and  subtract 
from  both  members  until  the  right-hand  member  is  zero. 
Factor  the  left-hand  member  by  inspection  if  you  can,  and 
solve  for  the  value  of  the  unknown  number.  If  you  cannot 
factor  readily,  write  the  values  of  the  unknown  by  means 
of  the  formula. 

If  a  geometric  figure  is  involved,  draw  a  careful  figure  for 
each  answer  if  possible.  If  irrational  numbers  are  involved, 
draw  by  making  use  of  the  Pythagorean  theorem. 

Always  examine  both  answers  to  see  if  each  has  a  mean- 
ing, or  if  one  is  to  be  discarded  as  meaningless. 

1.  The  altitude  of  a  parallelogram  is  7  units  less  than 
twice  its  base  and  the  area  is  10  square  units.  Find  the 
length  of  the  base  and  of  the  altitude  of  the  parallelogram. 

2.  The  upper  base  of  a  trapezoid  is  one  unit  less  than 

*  While  proceeding  with  the  solution  of  problems  in  this  list,  Parts  I 
and  II  of  chapter  VI  of  Geometry  may  be  studied. 


XIII.  §891  SUMMARY  — PROBLEMS  193 

one-half  the  lower  base.  The  altitude  is  5  units  less  than  1 J 
times  the  lower  base.  The  area  is  36  square  inches.  Find 
the  length  of  each  base  and  of  the  altitude. 

3.  In  the  adjacent  figure  we  have  a  circle  with  chords 
intersecting  at  point  0.  There  is  a 
theorem  which  states  that  the  product 
of  the  segments  of  one  of  the  chords 
is  equal  to  the  product  of  the  seg- 
ments of  the  other  chord.  By  meas- 
urement verify  the  truth  of  this 
statement. 

Supposing  that  OD  is  1  unit  less 
than  2  times  CO,  and  that  if  2  units  be  subtracted  from  COy 
it  will  be  3  times  as  long  as  OB,  while  AO  is  1  unit  more 
than  7  times  OB,  find  the  length  of  each  of  the  segments. 

4.  Making  use  of  segments  of  chords  intersecting  within 
a  circle  write  a  problem  leading  to  a  quadratic  equation. 

5.  Theorem.  If  from  the  vertex 
of  the  right  angle  of  a  right  tri- 
angle a  line  is  drawn  perpendicular 
to  the  hypotenuse,  the  square  of 
the  perpendicular  is  equal  to  the 
product  of  the  segments  into  which 
it  divides  the  hypotenuse.  Verify  this  by  measurement. 
In  the  right  triangle  ABC  with  CD  perpendicular  to  AB, 
suppose  that  CD  is  4  units  less  than  3  times  AD,  and  that 
DB  is  4  units  more  than  the  sum  oi  AD  and  CD.  Find  the 
lengths  of  the  lines.  Then  by  means  of  the  Pythagorean 
theorem  find  the  lengths  oi  AC  and  BC  and  check  by  show- 
ing that  AC^  +  CB^  =  AB\ 

6.  Making  use  of  the  theorem  just  stated  write  a  problem 
leading  to  a  quadratic  equation. 

On  p.  49,  it  was  brought  out  that  if  a  ball  was  hit  by  two 
mallets  the  result  of  the  action  would  be  the  same  whether 
the  strokes  were  simultaneous  or  the  second  stroke  was 


194  ALGEBRA  —  FIRST  COURSE  ixiii.  §  89 

given  when  the  ball  had  come  to  rest  after  the  first  stroke. 
This  is  true  even  if  the  strokes  are  given  at  right  angles  to 
one  another.  When  you  strike  the  ball  0  with  two  mallets  at 
right  angles  to  one  another,  one  having 
the  power  to  send  it  the  distance  0J5 
and  the  other  to  send  it  the  distance 
OA,  it  will  stop  at  the  point  C  no 
matter  whether  the  action  is  simulta- 
neous or  consecutive;  if  simultaneous, 
the  ball  will  go  along  the  path  OC;  if 
consecutive,  it  will  go  along  line  OB  to  B,  then  along  BC  to 
C.  So  that  it  is  possible  to  find  the  resultant  force  when 
two  forces  act  at  right  angles  to  one  another  by  making  use 
of  the  Pythagorean  theorem.  If  OA  equals  3  and  OB  equals 
4,  the  result  would  be  equivalent  to  that  of  a  blow  which 
would  send  the  ball  5  units  along  OC,  since  3^  +  42  =  5^. 

7.  The  resultant  of  two  forces  acting  at  right  angles  to 
one  another  is  10  pounds.  One  force  is  2  pounds  less  than 
the  other.     Find  the  number  of  pounds  in  each  force. 

Solution: 

Let  /  =  the  number  of  pounds  in  one  of  the  forces. 

Then         /  —  2  =  the  number  of  pounds  in  the  other  force. 

(/  —  2)2  +  /2  =  the  square  of  the  number  of  pounds  in  the  resultant 
force.     Why? 
But  100  =  the  square  of  the  number  of  pounds  in  the  resultant 

force. 
.-.       (/-2)2+/2  =  100. 
2/2  -  4/  +  4  =  100. 
2/2  -  4/  -  96  =  0. 
/2  -  2/  -  48  =  0. 
(/  +  6)  (/  -  8)  =  0. 
If  /  +  6  =  0, 

/  =  —  6,  the  number  of  pounds  in  one  of  the  forces, 
and  /  —  2  =  —  8,  the  number  of  pounds  in  the  other  force. 

If  /  -  8  =  0, 

/  =  8,  the  number  of  pounds  in  one  of  the  forces, 
and  /  —  2  =  6,  the  number  of  pounds  in  the  other  force. 


XIII,  §  891 


SUMMARY  —  PROBLEMS 


195 


Check  each  set  of  answers. 

The  drawing  for  this  is  shown  in  the  figure. 


8.  If  the  larger  of  two  forces  acting  at  right  angles  to  one 
another  is  5  lbs.  less  than  5  times  the  other,  and  the  resultant 
of  the  two  forces  is  4  less  than  5  times  the  other,  what  is  the 
magnitude  of  each  force  and  of  their  resultant  ?  Make  the 
drawing  in  each  case. 

9.  Two  forces  act  at  right  angles  to  one  another  with  a 
resultant  force  of  17  kilograms.  One  of  the  forces  is  1  kilo- 
gram less  than  2  times  the  other.     What  are  the  forces  ? 

10.  A  car  is  moving  at  the  rate  of  12  miles  an  hour.  What 
is  the  rate  of  motion,  with  reference  to  the  ground,  of  a  man 
crossing  the  car  at  the  rate  of  5  miles  an  hour? 

11.  The  speed  of  a  parcel  thrown  from  a  train  at  right 
angles  to  it  was  17  miles  more  than  the  speed  of  the  train. 
The  speed  at  which  it  went  was  3  miles  less  than  2  times  the 
speed  of  the  train.  What  was  the  speed  of  the  train?  Of 
the  parcel? 


196  ALGEBRA  —  FIRST  COURSE  [Xili,  §  89 

12.  *  Write  three  problems  on  forces  or  velocities  acting 
at  right  angles  to  one  another. 

13.  Two  men  can  do  a  piece  of  work  in  6  days.  How 
long  will  it  take  each  of  them  working  alone  to  do  the  work 
if  it  takes  one  of  them  5  days  longer  than  the  other? 

Hint.  It  will  be  necessary  to  find  the  amount  each  can 
do  in  one  day  in  order  to  form  the  equation. 

Let  t  =  the  number  of  days  that  it  takes  one  to  do 

the  work. 
Then    t -{•  b  =  the  number  of  days  that  it  takes  the  other 

to  do  the  work. 

-  =  the  part  of  the  work  that  the  first  can  do  in 
z 

one  day. 
.  ,    g  =  the  part  of  the  work  that  the  second  can 
do  in  one  day. 
-  +         -  =  the  part  of  the  work  that  both  can  do  in 
one  day. 
le  part  of 
one  day. 

Finish  the  solution  and  check. 

14.  A  can  reap  one  and  one-half  times  as  fast  as  B,  and  B 
can  reap  one  and  two-thirds  times  as  fast  as  C.  How  many 
hours  will  each  require  to  reap  a  field  of  grain  which  all  three 
together  reap  in  30  hours? 

15.  A  vessel  can  be  filled  by  one  of  its  two  pipes  in  two 
hours  less  time  than  by  the  other,  and  by  both  together  in 

*  To  get  rational  numbers  which  may  be  used  as  the  sides  of  a  right 
triangle  substitute  special  values  for  m  and  n  in  the  expressions 
m^-\-n^y  2mn,  m^  —  n^.    Why  is  this  true? 


But  ^  =  the  part  of  the  work  that  both  can  do  in 


XIII.  §89]  SUMMARY  — PROBLEMS  197 

2  hours  and  55  minutes.     How  long  will  it  take  each  pipe 
alone  to  fill  it? 

16.  Three  men  can  finish  a  job  hi  1  hour  and  20  minutes. 
Working  alone,  C  would  take  twice  as  long  as  A  and  2  hours 
longer  than  B.  How  long  would  it  take  each  one  to  do  it 
alone? 

17.  A  steamer  on  account  of  poor  coal  makes  3  miles  per 
hour  less  speed  than  usual  and  requires  165  hours  more 
time  to  make  a  trip  of  4840  miles.     What  is  the  usual  speed  ? 

18.  A  vessel  steams  at  the  rate  of  11 /a  miles  per  hour. 
It  takes  as  long  to  steam  23  miles  up  the  river  as  47  miles 
down.  Find  the  velocity  of  the  river  and  time  required  for 
given  distances. 

19.  A  sets  out  on  a  journey  at  the  rate  of  Va  miles  per 
hour,  and  m  hours  later  B  sets  out  after  him  at  the  rate  of 
Th  miles  per  hour.  In  how  many  hours  will  B  overtake  A 
and  how  many  miles  will  each  have  walked? 

Make  a  careful  drawing  in  each  of  the  following  and  com- 
pare result  with  answer  found  by  algebra: 

20.  Of  two  rectangles  i^i  and  R^  the  base  of  i^i  is  2  times 
that  of  J?2-  The  altitude  of  Ri  is  h  units  and  that  of  R^  is  a 
units.  If  4  6^  square  units  be  subtracted  from  the  area  of 
Ri  and  a^  square  units  be  subtracted  from  the  area  of  R2,  the 
product  of  the  areas  will  be  4  a^h'^.     Find  the  base  of  each. 

21.  In  a  right-angled  triangle  one  side  is  one  unit  less 
than  twice  the  other,  and  the  hypotenuse  is  17  units.  What 
is  the  area  of  the  triangle? 

22.  The  altitude  of  a  parallelogram  is  27  r  inches  less  than 
9  times  the  base,  and  the  area  is  22  r^  square  units.  What 
are  its  dimensions? 

23.  A  regular  octagon  (a  figure  with  eight  equal  sides 
and  eight  equal  angles)  is  formed  by  cutting  off  the  corners 
of  a  square  whose  side  is  1  foot.  Find  the  side  of  the  octa- 
gon. 

24.  If  it  takes  A  7  days  longer  than  it  takes  B  to  do  a 


198  ALGEBRA  —  FIRST  COURSE  ixill.  §  89 

piece  of  work,  and  they  both  together  take  8f  days  to  do  it, 
how  long  will  it  take  each  working  alone  to  do  it  ? 

25.  If  it  takes  A  a  days  longer  than^B  to  do  a  piece  of 
work,  and  they  both  working  together  take  t  days  to  do  it, 
how  long  will  it  take  each  working  alone  to  do  it  ? 

26.  Using  the  answers  to  Exercise  25  as  formula,  solve 
Exercise  24  by  substituting  the  special  values  there  given  in 
the  answers  of  Exercise  25.  Compare  results  with  those 
obtained  when  you  solved  Exercise  24. 

27.  Write  a  special  problem  of  your  own,  which  may  be 
solved  by  using  the  answers  to  Exercise  25  as  formula.    Solve. 

28.  Solve  your  problem  of  Exercise  27  without  using  the 
answers  of  Exercise  25.  Compare  results.  Compare  pro- 
cesses, and  explain  the  advantage  of  one  method  over  the  other. 

29.  If  one  side  of  a  rectangle  is  r  units  longer  than  n 
times  the  other,  and  the  area  is  a  square  units,  how  long 
and  how  wide  is  the  rectangle  ? 

30.  Write  a  special  problem  which  may  be  solved  by  using 
the  answers  to  Problem  29  as  formulae.     Solve. 

31.  Two  forces  are  acting  at  right  angles  to  one  another. 
The  larger  is  a  units  more  than  the  smaller,  and  the  resultant 
force  is  h  units  more  than  the  smaller.  Find  the  amount  of 
each  force. 

32.  Write  a  special  problem  which  may  be  solved  by 
using  answers  to  Problem  31^as  formulae.     Solve. 

33.  In  order  to  cover  a  table  top  which  contains  972 
square  inches,  it  was  found  that  a  strip  9  inches  wide  must 
be  cut  from  a  square  bought  for  the  purpose  in  order  that 
the  remainder  might  just  fit  the  table.  How  much  goods 
had  been  purchased? 

34.  In  making  a  centerpiece  a  lady  found  that  ^  of  the  length 
of  jthe  side  of  the  linen  square  out  of  which  she  cut  the  circu- 
lar center  must  be  1  inch  more  than  the  width  of  the  lace  for 
the  border.  She  had  160f  square  inches  of  lace.  What  was 
the  side  of  the  square  out  of  which  she  cut  the  center  ? 


XIII,  §  89] 


SUMMARY  —  PROBLEMS 


199 


Hint.    You  may  use  V-  for  tt  in  this  problem. 
36.   Write  a  problem  about  circles  which  will  lead  to  a 
quadratic  equation. 

36.  How  wide  would  a  rug  28 
inches  long  have  to  be  in  order  to 
have  a  border  of  2  inches  on  all 
sides  and  a  design  of  two  equilat- 
eral triangles  standing  one  on  each 
side  with  its  vertex  on  the  opposite 
side,  as  shown  in  the  drawing? 

37.  A  class  decorating  commit- 
tee was  instructed  to  buy  mate- 
rial of  two  colors  for  a  table  cover;  the  center  to  be  of 
one  color  and  the  border  of  the  other.  The  table  was  8 
inches  longer  than  wide.  They  found  material  for  the  center 
that  was  just  as  wide  as  the  table  was  long,  so  they  bought 
a  piece  as  long  as  the  table  was  wide.  In  the  other  color 
they  found  material  for  the  border  which  when  divided  into 
4  strips  would  give  a  width  which  if  the  center  were  2  inches 
wider  would  be  \  the  width  of  the  center.  They  decided  to 
purchase  a  length  of  this  equal  to  the  length  of  their  cover 
and  to  use  the  extra  384  inches  for  pennants.  The  center 
material  was  $1  per  yard,  and  the  other  was  85  cents  per 
yard.  What  was  the  cost  of  the  cover  ?  By  drawing  show 
the  most  economical  way  for  cutting  the  pennants  out  of 
the  material  left. 

38.  Write  a  problem  about  a  rectangular  flower  bed  with 
a  border,  the  solution  of  which  will  involve  a  quadratic 
equation. 

39.  The  first  of  two  automobiles  weighs  \  ton  less  than  the 
second.  The  number  of  miles  per  hour  that  the  second  is 
running  is  6  more  than  12  times  the  number  of  tons  in  the 
weight  of  the  second,  while  the  number  of  miles  per  hour  that 
the  first  is  running  is  3  more  than  11  times  the  number  of 
tons  in  the  weight  of  the  second.    The  momentum  of  the 


200  ALGEBRA  —  FIRST  COURSE  txill.§89 

second  is  15  less  than  twice  that  of  the  first.     Find  the 
weight  and  speed  of  each. 

40.  Two  boys  starting  from  camp  to  fish  at  a  point  7 
miles  away  found  themselves  short  of  flies.  They  decided 
that  one  should  ride  to  a  store  10  miles  distant  and  from 
there  walk  to  the  point,  a  distance  of  4  miles.  In  order  to 
reach  the  point  at  the  same  time  it  was  necessary  for  the  boy 
to  ride  at  a  rate  2  miles  more  per  hour  than  twice  the  rate 
that  the  other  boy  walked  and  to  walk  ^^  of  a  mile  per  hour 
more  than  the  other  boy  walked.  What  was  the  rate  each 
walked  ?     At  what  rate  did  the  boy  ride  ? 

41.  Of  two  sums  of  money  which  were  drawing  simple 
interest  one  was  $200  more  than  twice  the  other.  The  rate 
of  interest  on  the  smaller  sum  was  $2  more  than  twice  the 
rate  on  the  larger  sum.  The  interest  on  the  smaller  sum 
was  $40  and  the  interest  on  the  larger  sum  was  $24.  What 
were  the  amounts  invested?  What  was  the  rate  of  interest 
each  drew? 

42.  Two  spheres  were  to  be  decorated,  one  gilded  and  the 
other  painted.  The  cost  for  gilding  a  square  foot  was  | 
cent  more  than  3  times  the  cost  for  painting.  The  number 
of  square  feet  in  the  one  to  be  painted  was  12.4  more  than 
11  times  the  number  of  square  feet  in  the  one  to  be  gilded. 
The  price  paid  for  the  work  was  $18.60  for  painting  the 
larger  sphere  and  $5.58  for  gilding  the  smaller.  What  was 
the  price  per  square  foot  for  each  ?  What  was  the  radius  of 
each?     (Here  let  tt  be  3.1.) 

Exercises  and  Problems  for  Review. 

The  following  exercises  are  suggested  for  review,  to  be 
assigned  one  or  two  a  day  with  the  work  which  follows. 

Draw  rectangles  whose  areas  shall  represent  the  following 
algebraic  expressions;    then  write  the  factors  of  these  ex- 


4.   ah  +  ac  —  ad. 

2.   a^-2ab  +  h\ 

6.   a2  +  5a  +  6. 

3.  a^-hK 

6.   6a2-a-  12. 

XIII.  §89]  SUMMARY  — PROBLEMS  201 

Factor  the  following  expressions. 

7.  15  p2  +  5  p  -  5.  24.   a2  (6  -  1)  -  h'  {b-1). 

8.  2/2-8  2/  +  12.  25.   a^  +  b\ 

9.  a^  +  ac  +  a6  +  c6.  26.   a^  —  6^. 

10.  /c2  +  2  afc  +  3  6/c  +  6  ah.   27.  a^  -  6^  -  a  +  6. 

11.  9  -  6  r  +  7-2.  28.  8  A;3  +  /i3  _  2  /c  -  h. 

12.  9  m2?22  —  24  mn  +  16.       29.  r^  (m  +  n)  —  ar  (m  +  n). 

13.  ^2  _  9  ^s  -I-  20  s2.  30.  s^  -  r^  -  s  +  r. 

14.  m^  -2m-  15.  31.  (a  +  6  +  c)^  -  a^. 

15.  6  m^  +  mn  —  2  n^.  32.  a^  -\-  ac  -\-  ad  +  cd. 

16.  3/i2-10/i  +  3.  33.  a^ -Sa%  +  3ab^-bK 

17.  15  —  a;  —  6  x^.  34.  x^  —  5x'^  -{-  x  —  5. 

18.  dy  -by  -b-\-d.  35.  a;^  +  6  x^  +  4  a;  +  24. 

19.  r  (r  -  1)2  -  r  +  1.  36.  a2  -  c^  -  2  c(^  -  d"^, 

20.  ac  -  a  -  c  +  1.  37.  a^  -  6^  -  2  a2  +  2  62. 

21.  9  -  25.  38.  1  -  7-2  -  2  rs  -  s^. 

22.  {a  —  6)2  —  c2.  39.  r  —  r^  —  2  rh  —  rs^. 

23.  m^  -  {p-  qy.  40.  (a  +  (^)2  -  4  (a  +  d)  +  4. 

41.  a2  +  2  a6  +  62  —  m2  —  2  mn  —  n^, 

42.  (s  -  r)  (a2  +  62)  -  2  a6  (s  -  r). 

43.  p2  _  ^2  _j_  2  pg  _  ^2  ^  2  pi  -  p2. 

44.  a;2  -  9  62  +  9  2/2  -  a2  +  6  a6  -  6  rc2/. 

45.  49  x2  +  4  2/2  -  9  a2  -  16  62  -  28  xy  -  24  ab, 

46.  |c2d-H'.  53.    -2+— -— 2* 

n2      m?i      m2 

47.  |m3  +  8n3.  64.   3-  — +i^- 
**  a        a^ 

Q        ^0 

48.  a2-a6  +  i62.  55.   -^  +  —  +  25. 

X  X 

49.  ^a^-^b\  56.    (a +  6)2 +  5  (a +  6) +  6 

60.  i--3-  67.    3^2  +  81^  +  4. 

61.  —3  +  -3-  68.    (a  +  6)2  -  (a  -  6)2. 


m- 


^°  +  ?^-  69-    (x  +  2/p  -  (x  -  y)'. 


r       r 


202  ALGEBRA  —  FIRST  COURSE  [Xlii.  §  89 

60.  (a)    (a -6)2 -5  (a -6) +  6. 

(6)  3(m  +  n)2-3'(m  +  n)  -  18. 

(c)  4(p-g)2+14s(p-5)  +  6s2. 

(c^)  6s(p-5)2  +  2U2(p_5)_|_9s3. 

(e)  mna:^  +  {mp  +  ng)  a;  +  pq. 

61.  (a)  a2-c2-2cd-(^2^ 

(c)  r  —  r^  —  2  r^s  —  rs^. 

(d)  r2 -r^- 2r3s- rV. 

(e)  r"  —  r"+2  _  2  r^+^s  —  r"s^. 
(/)  r*""^  —  r'^^^  —  2  r"s  —  r"~^s2. 

62.  (a)  a2"  -  62"». 
(6)  a2"+i  -  062"^. 

(C)      a2«+2  _  (j252m^ 

(d)   a2"+'"  —  a'"62m^ 

63.  (a)    |^\  a"-2  -  ^  a"62»». 
(6)    fc26Z-H'. 

(c)    m^*-  — Gw  — 7. 

((f)      p2n+«  _  9  pn+.  _   10  p«. 

(e)      p2n+sg  _|_  pn+sqtn+l  _  2  p8q2m+l^ 

if)   a'-¥-2a'  +  2b\ 

64.  (a)   r2  +  5r  +  6-r-3. 

(6)      p2n  _  4  pn  _|_  4  _  pn  _|.  2. 
(C)      p2n_4pn_^4^p3n_8, 

66.   Solve  for  the  value  of  p  that  will  verify  the  equation 

3p-  16^5 
p  3* 

66.  A  river  flows  at  the  rate  of  2  miles  an  hour,  and  a 
fisherman  finds  that  he  can  row  upstream  a  few  miles  in  6 
hours,  but  that  it  takes  him  only  3  hours  to  come  back.  How 
fast  does  the  fisherman  row  in  still  water? 

67.  Two  balls,  one  of  lead  containing  3  cc.  and  the  other 
of  granite  containing  7  cc,  are  set  in  motion.    One  moves  at 


XIII,  §  89]  SUMMARY  —  PROBLEMS  203 

the  rate  of  14  cm.  in  4  sec,  and  the  other  at  the  rate  of  27 
cm.  m  5  sec.  The  momentum  of  the  lead  ball  is  19.11  units 
more  than  that  of  the  granite  ball.  If  the  density  of  lead 
is  .76  more  than  4  times  that  of  granite,  what  is  the  density 
of  each  ?     What  is  the  momentum  of  each  ? 

68.  A  and  B  start  from  Lincoln  in  the  direction  of  Omaha. 
At  noon  A  has  gone  f  of  the  distance  to  Omaha,  and  B  has 
gone  ^  of  the  distance,  and  they  are  just  If  miles  apart. 
What  is  the  distance  from  Lincoln  to  Omaha? 

69.  Factor: 

81r2-225;  1  -  100  hK 

^.      ,.«       x'^  -  X  -20  x^  -  x-2        x  +  1 
Simplify : 


a:2  -  25  2  a;  +  8  x^ -{- 5x 

70.  A  man  rows  9  miles  downstream  in  45  min. ;  he  rows 
back,  near  the  bank,  where  the  current  is  only  half  strength, 
in  li  hours.  What  is  the  speed  of  the  boat  and  of  the 
stream  ? 

71.  In  a  three-digit  number  the  tens  digit  exceeds  the 
hundreds  digit  by  3.  The  units  digit  is  4  less  than  2  times 
the  hundreds  digit.  Interchanging  the  units  and  tens  digit 
decreases  the  nimiber  by  45.     What  is  the  number? 

air  1    ,       2  13  5p 

72.  Solve  for  p:    p, -f  - 


2  '  p  +  2       8       4p  +  8 

73.  We  are  to  construct  a  square  and  two  rectangles 
such  that  the  first  rectangle  has  a  length  2  units  more  than 
and  J  less  than  the  side  of  the  square.  The  second  rectangle 
has  a  length  4  units  more  than  and  a  width  2f  units  less 
than  the  side  of  the  square.  The  areas  of  the  two  rectangles 
are  the  same.  Find  the  dimensions  of  each  of  the  figures 
and  make  the  drawing. 


204  ALGEBRA  —  FIRST  COURSE  [xiii,§89 

74.  Factor  the  following: 

a2  +  3a  +  2;  4  -  a^; 

m2  -  10  m  +  16;  SG^  -  U^; 

4p2  +  4  p  +  1;  (w  +  2)2  +  2  (m  +  2)  +  1; 

52  +  2.5  + 1;  z'^-Sz-^O. 

75.  If  P  is  any  point  within  the  parallelogram  ABCD, 
prove  that  the  triangle  PAB  plus,  the  triangle  PCD  equals 
one-half  of  the  parallelogram. 

76.  If  the  velocity  of  a  body  flying  through  space  is  144 
miles  a  second,  and  a  particle  is  thrown  from  it  at  right 
angles  to  its  course,  with  a  velocity  of  130  miles  a  sec, 
what  is  the  velocity  of  the  particle  at  the  moment  it  leaves 
the  body? 

77.  A  triangle  X  is  equal  to  a  fixed  triangle  T,  and  has  a 
common  base  with  T.  On  what  line  or  lines  must  the  vertex 
of  Z  fall? 

78.  Simplify:        x-7- -r  -  ^.      ,     ,  + 


2  {m  —  n)      2  (m  +  n)      m^  (m^  —  n^) 

79.  Two  forces  are  acting  at  right  angles  to  one  another. 
The  smaller  of  the  two  is  1  kilogram  less  than  the  larger, 
and  the  resultant  force  is  8  kilograms  more  than  the  larger. 
What  is  the  intensity  of  the  two  forces? 

80.  If  P  is  a  point  on  the  side  AB  oi  the  parallelogram 
ABCD,  and  Q  is  any  point  on  the  side  CD,  prove  that 
triangle  PCD  equals  triangle  QAB. 

81.  The  side  of  an  equilateral  triangle  is  s;  find  its  altitude 
and  then  its  area. 

82.  Use  the  answer  to  Exercise  81  as  a  formula  to  find  the 
following  areas :  s  =  8;  s  =  15;  s  =  10. 

1  Ci  d^  —  4 

83.  Simplify:  — r-:r 5 r— r  +  -T-T— r* 

^    "^     a  +  1      a^  —  a  +  1      a^  +  1 

84.  Of  two  rectangles  having  the  same  width,  the  length 
of  the  first  is  a  units,  and  of  the  second  is  h  units.  If  a 
square  of  side  a  units  be  subtracted  from  the  first  and  a 


XIII,  §  89]  SUMMARY  —  PROBLEMS  205 

square  of  side  b  units  be  subtracted  from  the  second,  the 
areas  of  the  figures  left  will  be  equal.  Find  the  width  of 
each  rectangle  and  draw. 

86.   Simplify:  ^+^  -  P^,- 

X  —  y      x^  —  y^ 

86.  A  train  is  going  at  the  rate  of  56  miles  an  hour.  A 
parcel  is  thrown  from  it  at  right  angles  to  the  direction  of 
the  train.  The  velocity  of  the  parcel  at  the  moment  it  leaves 
the  train  is  65  miles  an  hour.  With  what  velocity  was  the 
parcel  thrown? 

87.  Find  the  value  of  r  that  will  verify  the  following 
equation: 

r-  2      r-  1      2r  +  4 
r  +  2~r  +  l"^r2-r 

88.  Factor: 

s3  +  8H;  27a3-6463; 

St' -\- 27 a';  a'b'-8c'; 

m'  —  1. 

89.  A  lead  ball  moving  at  the  rate  of  7  cm.  per  sec.  stops 
an  ivory  ball  moving  at  the  rate  of  8  cm.  per  sec.  The  balls 
measure  respectively  13  cc.  and  71  cc.  If  1  cc.  of  lead 
w^eighs  6.8  grams  less  than  10  times  the  same  amount  of 
ivory,  what  is  the  density  of  each  substance? 

90.  Write  the  formula  for  the  area  of  a  trapezoid,  and 
solve  for  h.  Use  this  formula  to  find  the  altitude  of  a  trape- 
zoid whose  bases  are  34  units  and  45  units  respectively,  and 
whose  area  is  237  sq.  units. 

91.  Factor: 

9  a2  +  30  d  +  25;  a^-^2  aH  +  ad}) 

4m2-  12m +  9;  5r2-  lOr-f-20; 

a;2  -  a;  -  6;  2  m^  +  3  mn  -  2  n^; 

a;2  +  7rc-30;  2a2  +  a-15; 

r3-8s3;  3r2-  17r  +  10; 

.    27m3  + 64/^3;  (^_|-^)2_io(m  +  n)~39; 

6a;2  +  x-12;  9>t^  -  I, 


CHAPTER  XIV 

VARIABLES.     CONSTANTS.     FUNCTIONS  OF  VARI- 
ABLES.    GRAPHIC  REPRESENTATION  OF 
FUNCTION  OF  A  SINGLE  VARD^LE 

90.  Variables.  Up  to  this  time  in  your  work  algebra  has 
treated  almost  entirely  of  literal  arithmetic.  The  material 
has  been  furnished  from  several  sources,  principally  geome- 
try and  physics.  You  will  continue  to  use  quantity  fur- 
nished by  these  as  a  basis  for  your  discussions,  but  you 
will  now  study  quantities  in  their  relation  to  one  another. 
If  there  is  a  change  in  any  element  of  a  geometric  figure, 
there  are  consequent  changes  in  other  elements  which  de- 
pend upon  it.  If  there  is  a  change  in  any  magnitude  in  the 
physical  world,  there  must  be  consequent  changes  in  other 
magnitudes  which  depend  upon  it.  It  is  this  dependency 
of  quantity  upon  quantity  that  we  shall  now  investigate. 

Definition.  A  quantity  whose  magnitude  changes  is 
called  a  variable  quantity,  or  simply  a  variable. 

Examples  of  variables  are  found  everywhere  about  us;  we 
mention  a  few. 

The  height  of  a  growing  tree. 

The  weight  of  a  growing  apple. 

The  temperature  of  outdoor  air. 

The  barometric  pressure. 

The  distance  of  a  moving  train  from  a  station. 

The  speed  of  a  falling  body. 

The  volume  of  an  expanding  soap  bubble. 

Such  examples  may  be  multiplied  indefinitely. 

206 


XIV.  §92]  FUNCTIONS  207 

91.  Constants.  A  quantity  whose  magnitude  is  fixed  is 
called  a  constant  quantity,  or  simply  a  constant. 

In  nature  almost  everything  is  changing  or  variable,  so 
that  it  would  be  hard  to  find  anything  that  is  absolutely 
fixed  or  unchanging.     But  many  things  are  very  nearly  so. 

For  example  pick  out  any  object  about  you,  in  the  room 
or  out  of  doors.  What  can  you  say  as  to  its  size,  form, 
weight,  color? 

In  algebra,  then,  constants  are  quantities  which  for  the 
time  being,  or  in  connection  with  a  certain  problem,  we 
suppose  to  be  fixed  in  value. 

92.  Functions  of  Variables.  Many  variable  quantities 
depend  upon  one  or  more  other  variable  quantities.  We 
give  some  examples. 

Example  1.  Draw  two  lines  at  right  angles  to  each  other,  and  on 
them  mark  scales  on  which  to  count  the  bases  and  altitudes  of  rect- 
angles. Using  these  same  lines  and  counting  from  the  origin  each  time, 
draw  the  following  rectangles: 


Base  5  and  altitude  1. 
Base  5  and  altitude  2. 
Base  5  and  altitude  3. 
Base  5  and  altitude  4. 

Examine  carefully  these  rec- 
tangles. You  have  kept  the  base 
the  same  and  have  changed  the 
altitudes.  Did  anything  else  about 
your  rectangles  change?  Using 
the  customary  notation  to  repre- 
sent the  number  of  units  in  the 
area,  base  and  altitude  of  a  rectangle,  we  have 

a  =  bh. 

We  see  that  as  we  varied  the  altitude  h  and  allowed  the  base  h  to 
remain  constant,  the  value  of  the  area,  a,  changed.  That  is,  a  takes 
on  a  new  and  definite  value  to  correspond  to  each  new  and  definite 
value  that  we  give  to  h.  In  other  words,  a  depends  upon  h  for  its 
value  while  b  remains  constant. 


T — I — 1 — r 


208 


ALGEBRA  —  FIRST  COURSE 


[XIV,  §  92 


but 


Definition.  When  one  variable  depends  upon  another  for 
its  value,  we  say  that  the  first  is  a  function  of  the  second. 
When  a  quantity  depends  upon  several  others,  it  is  said  to 
be  a  function  of  these  several  quantities.  In  the  equation 
above,  to  express  the  thought  brought  out  by  our  drawing, 
we  say  that  a  is  a  function  of  h.  This  is  written: 
a  =  f  {h).     (a  is  a  function  of  h.) 

Both  h  and  a  are  variables,  and  to  distinguish  them,  we 
call  h  the  independent  variable,  and  a  the  dependent  variable. 
Make  a  drawing  similar  to  the  one  just  made,  but  con- 
sidering h  the  constant,  h  the  independent  variable,  a  the 
dependent  variable.     In  this  case,  the  equation  is  the  same, 

a  =  hh, 
a=f{h). 

Make  a  drawing  changing  both  the  base  and  altitude  at  the 
same  time.  Still  a  =  hh,  but  now  a  =  f  (b,  h),  that  is,  a  is 
a  function  of  b  and  h.  Here  a  depends  upon  both  b  and  h 
for  its  value. 

Explain  how  the  same  idea  may  be  illustrated  by  a  triangle; 
a  trapezoid. 

Example  2.  Draw  a  right-angled 
triangle,  calling  the  sides  x,  y,  r. 
Considering  y  as  constant,  using  the 
same  origin,  draw  triangles,  letting 
X  =  1,  2,  3,  4,  etc.  See  the  adjacent 
figure,  which  is  drawn  for  y  =  2. 

Examining  the  drawings  what  do 
you  find  as  to  the  length  of  r?  As 
you  have  learned  r^  =  x^  +  y^;  in  this  example  r^  =  x^  +  4.  There- 
fore r  =f{x). 

Consider  x  as  the  constant  and  y  as  the  independent  variable,  make 
a  drawing.     State  the  equation  of  functional  relation  of  the  variables. 
Make  drawings  when  both  x  and  y  vary.     Write  the  equation  of 
functional  relation  of  the  variables. 

Make  drawings  to  illustrate  x  =f  (r),  taking  y  =  I.    Again,  take 
y  =  2.    Again,  take  2/  =  3. 

Example  3.     Draw  circles  using  as  radii  1,  2,  3,  4,  etc.    What  is 


y 

/ 

^ 

^ 

^ 

■ 

X 

0 

1 

i 

J          i 

XIV,  §  921 


FUNCTIONS 


209 


the  area  of  each?     If  a  is  the  number  of  square  units  in  the  area,  and  r 
the  number  of  linear  units  in  the  radius,  then 
a  =  irr^.    Here  a  =  f  (r). 

As  usual  T  denotes  the  numerical  constant  whose  value  is  about  -V-; 
more  accurately,  x  =  3.1416  — . 

Example  4.  Draw  a  circle.  At  the  center  draw  angles  of  10°,  20", 
30°,  etc.  These  are  called  central  angles.  Pick  out  various  magni- 
tudes in  the  figure  which  change  because  the  central  angle  is  changed. 

Definition.  A  portion  of  the  circumference  of  a  circle  is  called  an  arc. 
The  portion  of  the  area  of  the  circle  inclosed  between  the  arms  of  a  central 
angle  and  the  arc  is  called  a  sector. 

If  s  stands  for  the  number  of  square  units  in  the  area  of  the  sector, 
and  a  for  the  number  of  degrees  in  the  angle,  without  writing  the  equa- 
tion, we  may  express  the  dependency  of  the  magnitudes  s  and  a  by 
writing  s  =  f  (a).    In  this  case  the  radius  of  the  circle  is  the  constant. 

If  c  stands  for  the  number  of  units  in  the  length  of  the  arc,  write  the 
statement  of  dependency  of  c  on  a. 

Mark  arcs  of  various  lengths  on  the  circumference.  Join  the  ends  to 
the  center,  thus  forming  angles.  In  this  case  what  is  the  independent 
variable?     The  dependent  variable?    State  this  by  means  of  symbols. 

Example  5.  Draw  a  circle  and  draw  a  line  across  it  terminated  at 
each  end  by  the  circumference.  Such  a  line  is  called  a  chord  of  the 
circle.  Draw  several  chords  parallel  to  the  one  you  have  drawn,  each 
one  nearer  to  the  center.     Thus: 


The  distance  from  the  center  to  the  chord  is  represented  by  the 
perpendicular  line  drawn  from  the  center  to  the  chord.  By  testing 
you  will  find  that  this  perpendicular  bisects  the  chord. 


210  ALGEBRA  —  FIRST  COURSE  [xiv.  §  92 

The  area  of  the  portion  of  the  circle  bounded  by  the  arc  and  its  chord 
is  called  a  segment  of  the  circle. 

In  the  drawing,  what  magnitudes  change  because  the  chord  is  drawn 
nearer  to  the  center?  Assign  letters  to  the  various  magnitudes,  and 
express  the  dependency  by  symbol  language.  Do  all  magnitudes  grow 
larger  because  the  magnitudes  upon  which  they  depend  grow  larger? 

Draw  the  radii  to  the  ends  of  the  chords.  Then  we  have  the  equa- 
tion 

where  c  stands  for  the  number  of  units  in  the  length  of  the  chord,  r  for 
the  number  of  units  in  the  radius,  and  d  for  the  number  of  units  from  the 
center  to  the  chord.     Explain  how  we  get  this  equation. 
In  our  drawing 

c=f{d). 

Compare  this  with  Example  2. 

Suppose  you  draw  chords  such  that  each  one  is  longer  than  the  one 
preceding.  What  can  you  say  about  the  distances  from  the  center? 
Express  this  algebraically. 

Examples  of  Functional  Relations  not  Geometrical. 
Example  1.     Let         d  =  the  number  of  units  of  distance. 
V  =  the  number  of  units  of  speed. 
t  =  the  number  of  units  of  time. 
Then  the  relation  is 

d  =  vt.        Here  d  =  f  (v,  I). 

The  last  equation  is  read:  "i^  is  a  function  of  v  and  of  t." 
Example  2.     Let  R  denote  the  reading  of  a  barometer  at  a  height  h 
above  sea  level.     When  a  barometer  is  carried  up  a  mountain,  it  can 
be  seen  that  the  reading  changes,  but  we  do  not  know  precisely  the 
law  connecting  R  and  h.     We  can  say 

R  =  f  (h).     (E  is  a  function  of  h.) 

But  we  do  not  know  just  what  /  {h)  is.  So  we  merely  indicate  that 
R  depends  on  h  by  writing  R  =  f  (h). 

Example  3.  The  speed  of  a  falling  stone  depends  on  the  time  since 
it  started  to  fall. 

Let         s  =  number  of  units  in  speed  at  any  time, 

t  =  number  of  units  of  time  since  stone  was  dropped. 

Then  we  know  from  experiment  that 

s  =  16«2.        Heres=/(0. 

This  is  at  least  a  very  close  approximation;  it  is  not  the  exact  value 
of  8  in  terms  of  t,  but  serves  for  most  purposes. 


XIV.  §  92]  FUNCTIONS  21 1 

Eximyle  4.   The  interest  on  a  principal  depends  on  the  amount  in- 
vested, the  rate  of  interest,  and  on  the  time  during  which  it  is  invested. 
Let        /  =  the  number  of  dollars  interest  (simple  or  compound). 
P  =  the  number  of  dollars  principal, 
r  =  the  yearly  rate  of  interest. 
t  =  the  number  of  years. 
Then     /  =  PrL     Here  I  =fiP,  r,  t). 
That  is,  Z  is  a  function  of  three  variables,  P,  r,  t. 
Example  5.     The  temperature  of  the  outer  air  at  a  given  place  on  a 
given  day  depends  on  the  hour  of  the  day. 

Let         T  =  number  of  degrees  of  temperature, 

t  =  number  representing  time  of  day,   counted  from  any 
particular  moment,  as  noon,  say. 
Then    T=fit). 
But  here  we  do  not  know  just  how  to  express  T  in  terms  of  t. 

Exercises.* 

In  the  following  exercises  give  the  exact  form  of  the 
function. 

1.  The  distance  passed  over  by  a  train  going  30  miles  an 
hour  for  t  hours.     Here  d  =  f  (t).     What  is  /  (t)  ? 

2.  The  cost  of  n  yards  of  cloth  at  $2  a  yard.  Here 
c=f{n).     What  is /(n)? 

3.  The  time  required  to  go  one  mile  at  the  rate  of  v  feet 
per  second.     Here  t  =  f  (v). 

4.  The  cost  of  building  a  cement  walk  I  feet  long  and  w 
feet  wide  at  14  cts.  a  square  foot.     Here  c  =  f  {l,w). 

5.  The  cost  of  excavating  a  cellar  x  feet  long,  y  feet  wide 
and  z  feet  deep.     Here  c  =  f  {x,  y,  z). 

6.  Cost  of  making  n  photographs,  if  the  cost  is  30  cts.  to 
make  the  negative  and  10  cts.  each  to  print,  mount,  and 
finish  the  pictures. 

In  the  following  first  express  the  required  quantity  as  a 
function  and  then  give  the  precise  form  of  the  function  when 
possible. 

7.  The  cost  of  using  3  electric  lights  at  1  ct.  each  per  hour. 

*  Chapter  IV  of  "  Geometry  "  may  weU  be  introduced  while  the  class 
is  working  these  exercises. 


212  ALGEBRA  —  FIRST  COURSE  [xiv.  §  94 

8.  The  price  of  a  quantity  of  lumber  at  9  cts.  per  board 
foot. 

9.  The  cost  of  paving  a  street  at  $2  a  square  yard. 

10.  The  amount  of  lumber  needed  to  make  a  box. 

11.  The  amount  of  a  given  principal  which  is  placed  at 
5%  simple  interest. 

12.  The  weight  of  a  solid  rectangular  block  of  material. 
(Depends  on  volume  and  density.) 

13.  The  length  of  an  iron  rail  which  expands  when  heated. 
(Note  the  open  space  between  the  adjoining  rails  of  a  track.) 

14.  The  volume  occupied  by  a  cubic  foot  of  water  which 
expands  when  heated.  (Hence  the  need  of  an  expansion 
tank  in  heating  a  house  by  hot  water.  What  is  the  object 
of  such  a  tank  ?) 

93.  Functions  of  a  Single  Variable.  The  area  of  a  rec- 
tangle is 

a  =  hh. 

Here  a  is  a  function  of  two  quantities  b  and  h,  either  of  which 
may  vary.  Suppose  we  keep  h  fixed  and  let  h  alone  vary. 
Then  a  is  a  function  of  the  single  variable  h.     We  can  write 

a=f{h), 
since  b  has  a  fixed  numerical  value. 

We  shall  now  consider  some  functions  of  a  single  variable; 
any  letter  may  be  used  to  indicate  the  quantity  which  is 
varying;  then  all  other  letters  contained  in  the  expression 
with  which  we  deal  will  stand  for  fixed  numbers. 

As  a  general  rule  we  use  the  first  few  letters  of  the  alpha- 
bet to  indicate  fixed  quantities  and  the  last  letters  to  indicate 
variables.  But  quite  often  the  initial  letter  of  a  word  is 
used  to  indicate  a  constant  or  a  variable  without  regard  to 
this  rule. 

94.  The  Function  Notation. 

Example.    Consider  the  quantity  x^  —  4  x  +  3. 
It  is  a  function  of  x,  so  we  may  write 

fix)  =x2-4x  +  3. 


XIV.  §  94]  FUNCTIONS  213 

Such  a  function  might  have  come  from  a  variety  of  sources.  It 
might  represent  the  area  of  a  rectangle  whose  dimensions  are  x  —  I 
and  X  -  3.     For  (x  -  1)  (x  -  3)  =  x2  -  4x  +  3. 

It  might  be  the  weight  of  x  —  1  cubic  feet  of  stone  which  weighs 
x  —  3  pounds  per  cubic  foot.  And  so  on.  Let  the  student  give  other 
illustrations. 

We  shall  now  use  the  following  notation: 
/  (1)  shall  mean  the  value  of  /  {x)  when  x  is  replaced  by  1. 
/  ( —  2)  shall  mean  the  value  of  /  (x)  when  x  is  replaced  by    —  2 
and  so  on.     Thus 

/(I)  =  (l)2-4(l)+3  =  l-4  +  3  =  0. 
/(-  2)  =  (-2)2-4(-2)  +3  =  4  +  8  +  3  =  15. 
/  (0)  =  (0)2  -4  (0) +3  =  0-0  +  3  =  3. 
f{a)  =  a2  -  4a  +  3. 
/(-6)  =  (_6)2_4(_?,)  +3  =  62^46 +  3. 

To  calculate  /  ( —  1)  •  /  (2)  we  would  proceed  as  follows: 
/(-I)  =  (-1)2 -4  (-1) +3  =  1+4  +  3  =  8 
/*  (2)  =22 -4. 2  +  3  =  4-8  +  3  =  -!. 
/(-I)./ (2)  =8.(-l)  =  -8. 

Exercises. 

1.  /  (x)  =  1  —  X.  Calculate  the  numerical  value  of  /  (x) 
when  X  =  —  3,  —  2,  —  1,  0,  1,  2,  3.  Arrange  your  results 
in  tabular  form. 

Thus:         a;  =-3,   -  2,     ,       3. 

fix)  =     4,        3,     ,  -2. 

2.  f  (x)  =  3  —  2  a;.  Calculate  /  (x)  when  x  =—  2,  —  1, 
0,  1,  2. 

Arrange  in  tabular  form  as  in  Exercise  1. 

3.  fix)  =4+  X.     Calculate  /  (x)  when  a;  =  -  4,  -  2,  0,  2,  4. 
4..  fix)  =2  X- 5,     Calculate  /  (4),  /  (2),  /  (0),  /  (-  2).. 

5.  fix)  =x\  Calculate/ (0), /(I),/ (2),/ (3),/ (1)./ (3), 
/(6)-/(2). 

6.  /  W  =  3  a;2  4-  2  X  -  1.  Calculate  /  (0),  /  (2),  /  (-  1), 
/(-2)./(3),/(l)^/(-3). 

7.  fix)=^x^-Zx.  Calculate /(-I), /(I),/ (-3),/ (3), 
2/(5) -^/(- 5). 


214  ALGEBRA  —  FIRST  COURSE  [XIV,  1 94 

8.  /(x)  =  (x^  +  l)K     Calculate  /  (J),  /  (-  J),  /  (1)  -/(J), 

Sometimes  we  have  to  deal  with  two  or  more  functions  in 
the  same  problem.  For  example  (x^  —  1)  (x^  -{-  1)  can  be 
regarded  as  the  product  of  the  two  functions  x^  —  1  and 
x^-\-  1. 

If  we  say  that  /  (x)  stands  for  the  function  x^  —  1,  we 
cannot  say  /  (x)  stands  for  x^  -{-  1  also.  We  would  say,  for 
example,  F  (x)  stands  for  x^  -{-  1,  using  the  capital  letter  to 
indicate  that  F  (x)  is  a  different  expression  from  /  (x). 
Often  Greek  letters  are  used;  thus,  to  indicate  a  number  of 
different  functions  of  a:,  we  might  use  the  symbols 

fix),    F{x),     4>{x),    iA(a;),  etc. 

These  are  read : 
The  /  function  of  x. 
The  capital  F  function  of  x. 
The  Phi  function  of  x,  i 
The  Psi  function  of  x,  etc. 

Exercises.*  {Note.  The  symbol  =  is  often  used  in  pUce 
of  the  words  '' stands  for,"  or  "is  identical  with.") 

find  the  value  of  /  (x)  -  F  (x)  -  <!>  {x), 

Tr  r/N_>-^-6  ,  4 1_ 

-^W  V^;-^3_|.8-^5y.2_  iOr  +  20      r  +  2' 
solve  the  equation  /  (r)  =  0. 

*  At  this  point  it  would  be  well  to  take  up  the  study  of  the 
Trigonometric  Functions  in  Part  III,  Chapter  VI  of  Geometry,  up  to 
the  point  there  indicated. 


XIV.  §95]  GRAPHIC  REPRESENTATION  215 

3.   If  Jfe  y,z)  =    .^     <,    F(x,y,z)  = 


z{x-yy        ^"-'^'"^      x{y-xy 
find  the  value  oi  f  {x,  y,z)  —  F  (x,  y,  z). 

zx 
find  the  value  oi  f  {x,  y,  z)  -{-  F  {x,  y,  z)  -{-  <j>  {x,  y,  z). 
6.   If  f(r,  s)  =  ~  -  r,  F{r,  s)  ^- \-, 

find  the  value  of  ^y^- 

F{r,s) 

6.   If  f{x)  =  a;2  -  4a;  +  3,    F  (x)  =  x^  -  lOa:^  +  9, 

1  f  (x) 

write  the  value  of 


F{x) 
7.   If  /  (r)  =  r^  -  6  r2  +  5  r,  F  (r)  =  r2  +  2  r  -  35, 

write  the  value  o(~~t^- 
Fir) 

find  the  value  of  f  (x)  +  F  (x)  +  0  (a;). 

find  the  value  of/  (u)  '  F  (u). 

95.   Graphic  Representation. 

Example.     Given  the  function  f  (x)  =  S  —  2  x. 
Make  a  table  as  in  Exercise  1,  §  94,  showing  the  values  of  this 
function  for  a  number  of  assumed  values  of  x,  thus: 

x=  -2,     -  1,     0,     1,         2,         3,  etc. 
f{x)=      7,         5,     3,     1,     -1,     -3,  etc. 


216 


ALGEBRA  —  FIRST  COURSE 


tXIV,  §  95 


Usually  it  is  better  to  write  these  in  columns,  be-  —  2 

ginning  with  the  negative  values,  in  this  case  x  =  —  2  —  1 

and  going  step  by  step  to  a;  =  3;    this  gives  the  ad-  0 

jacent  column  of  values.  -j-  i 

An  inspection  of  these  columns  shows  that  as  x  +2 

increases  from  —  2  to  3,  /  (x)  decreases  from  7  to  —  3.  +3 


/(x) 

7 
5 
3 
1 
-  1 
-3 


We  now  give  a  scheme  for 
representing  these  values  by 
a  diagram  from  which  we  can 
see  at  a  glance  how  /  (x) 
changes  as  x  changes. 

Draw  two  lines  at  right 
angles.  On  one  of  them,  say 
the  one  running  from  left  to 
right,  mark  off  a  number 
scale  to  show  values  of  x,  aa 
in  the  figure;  the  zero  of  the 
scale  is  at  the  intersection  of 
the  two  hnes,  and  positive 
numbers  are  usually  marked 
off  to  the  right. 

At  —  2  on  the  x-scale,  or 
on  the  X-axis  as  it  is  usually 
called,  lay  off  a  perpendicular 
upward  7  units  long.  This 
indicates  the  value  of  /  (x) 
when  X  =  —  2. 

At  —  1  lay  off  a  perpen- 
dicular 5  units  long,  showing 
the  value  of  /  (x)  when  x 
=  —  1;  at  0  and  1  lay  off 
perpendiculars  upward  3  and 
1  units  long  respectively. 
The  perpendiculars  are  getting  shorter,  showing  that  the  values  of/  (x) 
are  decreasing. 

At  X  =  +  2,  /  (x)  =  —  1,  from  our  table.  To  show  this  on  our  dia- 
gram at  2  on  the  x-axis  draw  a  perpendicular  downward  one  unit  long. 
We  shall  understand  that  when  the  perpendicular  is  drawn  downward 
the  function  has  a  negative  value.  Similarly  when  x  =  3,  /  (x)  =  —  3; 
hence  at  3  on  the  x-axis  draw  a  perpendicular  downward  3  units  long. 


/( 

> 

-2X 

7 

e 

5 

u 

3 

' 

J 

\ 

~~ 

-' 

-0 

I 

2 

3 

-J 

-2 

-3 

XIV.  §  95] 


GRAPHIC  REPRESENTATION 


217 


It  is  now  evident  on  inspection  that  the  free  ends  of  these  perpen- 
diculars all  lie  on  a  straight  line.     In  the  following  figure  these  free 
ends    are    marked   by    heavy- 
dots    and    the    straight    line 
through  them  is  drawn. 


Exercise.  On  a  sheet  of 
cross-ruled  paper  draw  ac- 
curately the  above  figure, 
showing  the  perpendic- 
ulars, the  dots  and  the 
straight  line.  Now  draw 
in  some  more  perpendicu- 
lars showing  the  values  of 
3  —  2  a;  for  other  values  of 
Xj  such  as  a;  =  —  f ,  —  i, 

3      Ol 

Where  do  the  ends  of 
these  perpendiculars  fall? 
What  does  the  straight 
line  tell  you  about  the 
function  3  —  2  x  ? 

From  your  diagram  read 
off  the  values  of  3  —  2  a; 
when  a;  =-  If,  -  |,  i  IJ, 
2f .  Check  these  by  direct 
calculation. 

State  a  rule  for  finding  the  value  of  3  —  2  a;  for  any 
assumed  value  of  x  from  the  diagram. 

Definition.  The  straight  line  drawn  through  the  dots 
in  the  above  diagram  is  called  the  graph  of  the  function 
3  -2a;. 

This  graph  is  a  geometric  picture  which  shows  us  the  value 
of  3  —  2  a;  for  the  values  of  x  coming  within  the  limit  of  the 
diagram.  For  larger  values  of  x,  to  the  left  or  right,  we 
would  have  to  use  a  larger  paper,  or  a  smaller  scale. 


218  ALGEBRA  —  FIRST  COURSE  [xiv.  §  96 

Exercises.     Draw  graphs  showing  values  of  the  following : 

1.  3  +  2  a;,  from  x  =  -  3  to  a;  =  2. 

2.  2  a;  —  3,  from  x  =  —  2  to  x  =  4:. 

3.  2  —  \x,  from  x  =  —  2  to  a;  =  6. 

Graph  of  f  (x)  when  f  (x)  =  ax  -{-  h.  It  will  be  observed 
that  in  all  of  the  above  exercises  the  graph  is  a  straight  line. 

We  would  infer  from  this,  that  whenever  the  function  f  (x) 
has  the  form  ax  +  h,  its  graph  is  a  straight  line.  For  this 
reason  ax  +  6  is  called  a  linear  function  of  x. 

We  shall  assume  the  correctness  of  this  rule,  without 
stopping  to  give  a  complete  proof.  Can  you  give  such  a 
proof? 

Exercises.  Using  the  rule  just  stated  draw  the  graphs  of 
the  following.  Notice  that  you  will  need  only  two  points 
to  fix  the  entire  graph.  Hence  only  two  values  of  the  func- 
tion need  be  calculated. 


1.   x-1. 

3.   2  X  -  5. 

5.   3  +  ix. 

2.   x  +  2. 

4.    1  -  2  X. 

6.   2  -  3  X. 

In  place  of  the  letter  x  any  other  letter  might  be  used. 
Draw  the  graph  of  the  following  functions: 

7.  fit)  =<  +  3.  10.  /   (y)  =5-32/. 

8.  fir)  =2r-l.  11.  /    ih)  =ih  +  6. 

9.  /  to)  =  8  -  6  g.  12.   F  iw)  =  i  (^  -  2). 

96.   Graph  of  Quadratic  Functions. 
Definition.    A  function  whose  form  is 
/  (x)  =  ax^  +  6x  +  c 

is  called  a  quadratic  function  of  x. 

Here  a,  h,  c  are  given  numbers,  and  either  h  or  c,  or  both, 
may  be  zero;  but  a  must  not  be  zero,  for  then  our  function 
would  be  linear. 


XIV,  §  96] 


GRAPHIC  REPRESENTATION 


219 


The  graph  of  such  a  function  is  obtained  precisely  as  in 
the  case  of  those  just  studied.  It  is  not  a  straight  line  but 
has  nevertheless  a  very  simple  form. 

Example  1.    /  (x)  =  Ix^.     Here  a  =  \,  6  =  0,  c  =  0. 

We  first  make  our  table  of  values  of  /  (x)  for  assumed  values  of  x. 


X 

fix) 

-4 

4 

-3 

21 

-2 

1 

-1 

i 

0 

0 

1 

i 

2 

1 

3 

n 

4 

4 

Notice 

that,   on   ac- 

count 

of  the  x^,   we 

get   the   same   value 

for  /  (x)  when  x  =  —  4  and  when  x  =  +  4;    similarly  when  x  =  db  3, 

db  2,  etc. 

We  now  represent  these  values  of  /  (x)  by  perpendiculars,  as  before. 
The  free  ends  of  these  perpendiculars  we  mark  by  small  circles.  Then 
we  draw  a  smooth  curve  through  the  circles  and  so  we  have  the  graph 
qf  the  function  I  x^. 

If  we  take  any  value  of  x,  as  x  =  2|,  calculate  from  it  the  value  of 
\  x^,  and  draw  a  perpendicular  of  this  length  at  the  point  x  =  2^  on 
the  X-axis,  the  end  of  this  perpendicular  marks  another  point  on  the 
curve.  So  we  might  fill  in  any  number  of  points  besides  those  already 
shown.  As  a  rule  we  draw  only  enough  perpendiculars  to  clearly 
outUne  the  curve. 

From  the  diagram,  read  off  the  values  of  i  x^ 


Exercise. 

when  X  =—  Si,  —  IJ,  2|.     Check  these  by  direct  calcula- 
tion.    What  does  the  curve  tell  you  about  the  function  J  a;^? 

From  the  diagram  estimate  the  values  of  x  for  which 

1  ^2   _   O     1     Ol 

Example  2.    f  {x)  =  i  x^  +  2.     Here  a  =  i  b  =  0,  c  =  2. 
Notice  that  this  function  is  just  two  units  greater  than  the  function 
in  Example  1. 


220 


ALGEBRA  —  FIRST  COURSE 


[XIV,  §  96 


Making  the 

X 

table: 

-4 

6 

-3 
-2 

4i 
3 

-1 
0 

2i 
2 

1 
2 

2i 
3 

3 

4 

4i 
6 

Drawing    the    graph 

as  before  we  see  that  we 

^x  have   exactly  the  same 

curve,    only    raised    up 

two  units. 


Example  3.    /  (x)  =  i  x^  -  §  x  +  2.     Here  a  =  i,  b 


I,  c  =  2. 


/ 

V 

\ 

k 

/- 

\ 

/ 
/ 

\ 

GV 

/ 
/ 
/ 

\ 

\ 

5 

\ 

S" 

i 

- 

^ 

n 

r 

Y 

V- 

- 

'  0 

'a; 

x: 

-4 

3      - 

-2 

I 

0 

1      1      1 

1          2        3 

4 

/(a 

5): 

8 

5i 

4 

21 

2 

li 

2 

< 

2f 

4 

XIV,  §961  GRAPHIC  REPRESENTATION  221 

The  above  figure  shows  the  graph.    Suppose  we  extend  the  figure 
farther  to  the  right;  x  =  5,  f  {x)  —  ^^-;  x  =  6,  /  (x)  =  8. 
This  is  shown  by  the  dotted  part  of  the  figure. 

Exercise  1.  We  can  easily  verify  now  that  this  is  the 
same  curve  as  before.  Draw  these  three  curves  accurately 
on  separate  sheets  of  thin  paper.  Superpose  the  curves 
and  see  if  you  can  make  any  one  of  them  cover  up  either  of 
the  others. 

Exercise  2.  From  the  last  graph  estimate  the  value  of 
I  x^  —  i  X  -\-  2,  when  a:  =  —  SJ,  —  1  J,  4|.  Check  by  direct 
calculation. 

Also  estimate  the  value  of  x  for  which  J  a:^  —  J  x  +  2  has 
the  value  7;  5;  3. 

Definition,  The  curve  in  the  above  diagram  is  called  a 
parabola.  Such  curves  are  of  frequent  occurrence.  When 
a  telephone  wire  sags  in  the  middle,  its  form  is  that  of  a 
nearly  fiat  parabola.  Skyrockets  or  projectiles  from  guns  fol- 
low parabolic  curves.  Most  comets,  as  they  sweep  through 
the  solar  system,  describe  huge  parabolas. 

We  shall  ask  the  student  to  verify,  in  working  the  exercises 
below,  that  the  graph  of  a  quadratic  function  is  always  a 
parabola;  that  quadratic  functions  having  the  same  values 
of  a  but  different  values  of  b  and  c,  give  the  same  parabola, 
but  in  different  positions;  that  the  parabola  points  down- 
ward when  a  is  a  positive  number;  upward,  when  a  is  a 
negative  number.     The  curve  is  symmetrical  with  respect  to 

a  line  parallel  to  the  function  axis  and  distant—  pr-  from  it. 

2a 

Exercises.     Draw  the  graph  of  the  following  functions; 

superpose  the  curves  in  each  exercise  and  show  that  they  can 

be  made  to  coincide. 

1.   ix^;  ix2  +  l;  ^x^-x  +  1. 

l/y2.      1     l/y.2.      1     /v.    l/y.2 

3.  2x"-;  2x^  -  4:x;  2x^  -  4:X  +  2. 

4.  -2x^',  -2x^  +  4:x;-2x^  +  4:X-2. 


222 


ALGEBRA  —  FIRST  COURSE 


[XIV,  §  97 


5.  x2;  x^-lx',  a;2-  7x4-10. 

6.  -a;2;  -x'^-\-lx',  -x2  +  7aj-10. 


/(«!)  = 

.^. 

8x 

97. 

Graphs     of 

Other 

/ 

21 

Functions.     By    following 

1       step   by   step   the   process 

18 

5"     shown    in    the    preceding 

15 

e 

xamples,  we  can  draw  the 
raphs  of  many  other  func- 

g 

1"J 

9 

f/ 

illustrations. 

6 

^/ 

ff^\   ^ 

5+3. T- 

-1 

J(a5)=- 

10 

... 

3 

) 

^ 

/ 

IN 

0 

1     ) 

2 

3 

3 

S 

—  s 

-2 

r 

K 

Y 

^  ' 

d} 

j 

/ 

3 

1 

I 

Y 

/ 

" 

6 

. 

3      - 

2      - 

1      , 

^ 

Y 

^ 

/ 

~ 

9 

/ 

M 

'0 

1 

"' 

3"^ 

/ 

n 

/ 

Y 

1 

~ 

15 

/ 

2 

I 

— 

IS 

J 

\ 

5 

-i 

- 

til 

4 , 

' 

Example  1.    f{x)  =x^-S  x.          Example  2.    /  (x)  -          ^^ 

Table  of  Values.                                   Table  of  Values. 

X              fix)                                                          X              fix) 

-3-18                                         -3-3.7 

-2-2                                         -2        -1.5 

_1               2                                         -1        -0.5 

0               0                                             0            0.1 

1-2                                             1            0.3 

2              2                                            2            1.3 

8 

] 

L8 

3 

3.5 

XIV.  §  98]  SUMMARY  223 

Exercises. 

1.  From  the  figure  under  Example  1  estimate  the  following 
values: 

(a)  What  is  the  value  of  a;^  —  3  a;  when  x  =  |  ?  When 
a:  =  1.5?     When  x  =  2.75?     When  x  =-  2.25? 

(6)  For  what  values  oi  xis  x^  —  3  x  equal  to  zero  ?  Equal 
to  1?     Equal  to  -  1?     Equal  to  10? 

2.  From  the  figure  under  Example  2  estimate  the  following 
values: 

(a)  What  is  the  value  of  y^ when  x  =  1.5? 

When  X  =  2i?    When  x  =-  2-75? 

x^  -\-  S  X 1 

(6)  For  what  values  of  x  is  ^7] equal  to  1? 

Equal  to  -  1  ?    Equal  to  3  ?    Equal  to  -  3  ?    Equal  to  0  ? 

3.  Explain  how  you  would  obtain  an  approximate  value 
of  x  which  satisfies  the  equation  x^  -\-  S  x  —  1  =  0, 

4.  Draw  graphs  of  x^;  of  -^;  of  1  —  x^. 

o 

5.  Draw  the  graph  of  the  function 

f(x)=x^-3x^-2x-\-Q. 

From  your  figure  read  off  as  exactly  as  possible  the  solutions 
of  the  equation 

x^  -  3  x2  -  2  a;  +  6  =  0. 

6.  Solve  graphically  the  equation  2x^  —  5x  -{-  1  =0. 

98.   Summary. 

Definition.  A  variable  quantity  is  one  which,  in  a  given 
problem,  may  take  on  different  values. 

A  constant  quantity  is  one  which,  in  a  given  problem,  is 
supposed  to  be  fixed  in  value. 

There  are  two  kinds  of  variables;  independent  and  de- 
pendent. 

The  independent  variable  may  be  assigned  values  at  will. 


224  ALGEBRA  — FIRST  COURSE  [XIV,  §98 

The  dependent  variable  takes  its  value  from  the  inde- 
pendent variable. 

A  function  of  a  variable  is  a  quantity  whose  value  depends 
on  the  value  of  the  variable.  A  function  is  a  dependent 
variable. 

To  indicate  that  a  quantity  depends  on  a  variable,  say  Xj 
that  quantity  is  represented  by  the  symbol  /  (x),  which  is 
read  "function  of  x.^'  Then  the  symbol  f  (a)  means  the 
value  of  the  function  when  x  equals  a. 

A  function  of  a  variable  may  be  represented  graphically 
by  a  diagram  in  which  one  scale  shows  the  values  of  the 
variable  and  the  other  scale  shows  the  corresponding  values 
of  the  function. 

When  /  (x)  has  the  form  of  ax  +  h,  its  graph  is  a  straight 
line.     For  this  reason  ax  -\-  b  is  called  a  linear  expression. 

When  /  (x)  has  the  form  ax^  +  6a;  +  c,  its  graph  is  a 
parabola. 


CHAPTER  XV 


USES    OF    THE    GRAPH 

99.   Graphic  Representation  of  Measurements  of  a  Va- 
riable Quantity. 

Example  1.    Population  Curve.    From  the  Census  Reports  we  find 
the  population  of  the  United  States  as  follows: 


Year 

Population 

Year 

Population 

1800 

4.3  million 

1860 

31.4  million 

1810 

7 . 2  million 

1870 

38 . 6  million 

1820 

9.6  million 

1880 

50.2  million 

1830 

12.9  million 

1890 

62.6  million 

1840 

17.1  million 

1900 

76.3  million 

1850 

23.2  million 

1910 

92.2  million 

30  UO  50  60  70 

Population  Curve  for  U.S. 

225 


226 


ALGEBRA  — FIRST  COURSE 


[XV.  §  99 


By  inspection  of  the  figures  we  see  that  there  has  been  a  steady  rise 
in  population.  But  a  graphic  representation  of  the  data,  as  in  the 
above  figure,  brings  this  out  more  clearly. 

The  height  of  each  dot  above  the  line  running  from  right  to  left 
represents  the  population  for  the  corresponding  census  year;  by  draw- 
ing a  smooth  curve  through  the  dots  we  get  the  "population  curve." 
In  drawing  this  curve  we  assume  that  the  population  changes  gradually 
during  each  decade. 

Exercise.  From  this  diagram  estimate  the  population  in 
1845.     In  1855.     In  1868.     In  1883. 

Notice  that  the  gain  in  population  during  any  decade  is 
greater  than  during  the  preceding  decade.  Can  you  explain 
why?    Was  this  true  from  1860  to  1870?    Why  not? 

Example  2.  Temperature  Curve.  From  the  records  of  the  U.  S. 
Weather  Bureau'at  Lincoln,  Neb.,  we  take  the  following  temperatures 
for  April  26,  1910. 

Time  Temperature  Time         Temperature 

2  a.m.  +42°F.                              2  p.m.  +74°F. 

4  a.m.  40"  F.                              4  p.m.  77°  F. 

6  a.m.  40°  F.                              6  p.m.  70°  F. 

8  a.m.  52°  F.                             8  p.m.  67°  F. 

10  a.m.  63°  F.  10  p.m.  58°  F. 

Noon  69°  F.  Midnight  52°  F. 

From  these  readings  we  construct  the  following  figure.  Notice  that 
we  have  to  represent  temperatures  from  40°  to  77°.  We  therefore 
start  our  degree  scale  at  40°  in  place  of  at  0°.  In  drawing  a  smooth 
curve  through  the  dots  we  assume  that  the  temperature  changes 
gradually  from  one  value  to  the  next. 


1 

. 

y^T^ 

ir\° 

,x 

NJ 

k 

rrP 

I 

h 

K     ' 

1 

k 

rn° 

j. 

V 

r 

s 

'A 

. 

K 

n 

\  r^ 

r 

J     1 

0 

1 

2 

f      I 

»             ( 

5 

!         1 

0         1 

3 

A.M. 


Noon 
Temperature  Curve 


Midn. 


XV,  §  99] 


USES  OF  THE  GRAPH 


227 


Exercise.  From  the  temperature  curve  estimate  the 
temperature  at  7  a.m.  At  9.30  a.m.  At  5.30  p.m.  At  9 
P.M.  What  were  the  highest  and  lowest  temperatures  during 
the  day?    When  did  these  occur? 

Example  3.     Solubility    Curve.     By  experiment  we  can   find    the 
number  of  grams  of  common  salt  that  can  be  dissolved  in  100  grams  of 
water  when  the  water 
is  heated  to  different  S 

temperatures.    So  we         \s, 
get  the  following 
data,  from  which  the 
solubiUty     curve     is 
constructed. 


Temperature 
of  Water 

Weight  of 
Salt 

40°  C. 

36.6  gms. 

"8 
1? 

60°  C. 

37.2  gms. 

80°  C. 

38.2  gms. 

o 

100°  C. 

39.6  gms. 

•s 

^T 


Exercise.  From  Z 
the  diagram  read  e 
off  the  number  of 
grams  of  salt  that 
can  be  dissolved 
in  100  gms.  water 
heated  to  50°  C. 
To  85°  C.  To 
97°  C.  Find  the 
amount     of     salt 

that  can  be  dissolved  in  500  gms.  water  heated  to  65°  C. 
Find  the  number  of  ounces  of  salt  that  can  be  dissolved  in  a 
pint  of  water  heated  to  150°  F. 

Exercises.     Draw  curves  showing  the  following  sets  of  data. 

1.   Enrollment  in  public  schools  of  U.  S. 

Year:  1880  1890  1900  1910 

Pupils:  9.8  million   12.7  million   15.5  million    17.8  million 


Solubility  Curve 


228 


ALGEBRA  —  FIRST  COURSE 


[XV,  §  99 


2.   World's  production  of  gold. 


Year: 


1885  1890  1895  1900  1905  1910 


Gold  ounces:) 
(millions)     ) 

9.8       12.3      18.9      22.0 

.   Yield  of  wheat  per         4. 

Amount  of  one  dollar  at 

acre. 

5%  compound  interest. 

Year                  Bushels 

Years                   Amount 

1890            11.1 

10              $1.63 

1895            13.7 

20                2.65 

1900            12.3 

30                4.32 

1905            14.5 

40                7.04 

1910            13.9 

50              11.47 

Solubility  of  potassium 
carbonate  in  100  gms. 
water. 


Temperature 

20°  C. 
40°  C. 
60°  C. 
80°  C. 
100°  C. 


Salt  dissolved 

112  gms. 
117  gms. 
127  gms. 
140  gms. 
156  gms. 


6.  Boiling  point  of  water 
in  which  salt  is  dis- 
solved. 

_  Salt  dissolved 
in  100  gms.  water 

7  gms. 
12  gms. 


Boiling 
point 

101°  c. 


102°  C. 
104°  C. 
106°  C. 
109°  C. 


Average  height 
of  boys. 


22  gms. 
30  gms. 
40  gms. 

8.  Number  of  years  that 
persons  of  different  ages 
will  probably  live. 


Age 

6  years 
8  years 
10  years 
12  years 
14  years 
16  years 
18  years 
20  years 


Height 

44.0  inches 
47.0  inches 
51.8  inches 
55.0  inches 
59.3  inches 
64.3  inches 
67.0  inches 
67.5  inches 


Age 

0  years 
10  years 
20  years 
30  years 
40  years 
50  years 
60  years 


Expectation 
of  Life 

41  years 
47  years 
40  years 
33  years 
27  years 
20  years 
14  years 


Let  the  student  look  up  other  data  from  which  curves  may 
be  constructed. 
Study  the  following  curves  and  state  what  they  show. 


XV,  §  99] 


USES  OF  THE  GRAPH 


229 


Curve  showing  the  temperature  at  which  water  boils  when  various 
amounts  of  potassium  carbonate  are  dissolved  in  100  c.c. 


^ 

\ 

o 

^ 

Jr 

ISO 

^ 

^ 

-^ 

■s,.," 

/-. 

^^ 

■I'" 

-^ 

^ 

-^ 

t 

,^ 

cr 

^110- 

^^ 

-^ 

^ 
^ 

100°, 

o^ 

-T 

K 

■~n 

'^ 

^m. 


W 


m 


iO  100  120  IkO 

Grams  salt  dissolved 


IGO 


ISO 


200 


8  P.M. 


7  PM. 


6  P.M.- 


5  PM. 


UP.Mr 


SU- 
SS 

Curve  showmg  the  "5 
number   of    deaths  "^^^ 
from    tuberculosis  ?-^ 
per    10,000    popu-  ^2U 
lation  s^g 


Year 


Cwve  showmg  Local  Mecm  Ttma  of  SwhSjEi  in  Lctiiiajtde.  UO" 


*v 

/ 

" 

\ 

\ 

/ 

\ 

s 

/ 

\ 

/ 

/ 

\ 

V 

/ 

/ 

\ 

• 

/ 

\ 

I  I 


I  i 


Thermograph. 

Curve  showing  the  temperature  in  degrees  F.     (From  the  records  of  the 

U.  S.  Weather  Bureau  at  Lincoln,  Neb.) 

May  10.  May  11,  1910. 


Barograph. 

Curve  showing  barometric  pressure  in  inches.     (From  records  of  U.  S. 

Weather  Bureau  at  Lincoln,  Neb.) 

.Mar.  5.  Mar.  6,  1910.  Mar.  7. 


8  10X1124  68  10M246   8  10  XII  2468  10  M24  6  8  10  XII  2 


XV,  §  100] 


USES  OF  THE  GRAPH 


231 


100.   Graphic  Solution  of  Equations. 

Example  1.  Suppose  f  (x)  =  2  x  —  S.  For  what  value  of  a;  is 
/  (x)  equal  to  zero  ? 

Graphic  solution:  Draw  the  graph  of  the  function  2  x  —  3.  It  is 
a  straight  Une,  so  only  two  points  are  needed. 

x  =  0,    fix)  =  -3; 
x  =  S,    f{x)=  3. 

So  we  get  the  graph  as  shown  below. 

Reading  off  the  value  of  x  where  the  graph  crosses  the  x-axis,  we 
get  x  =  |.     This  is  the  required  solution.     Explain  why. 


■-X 


Check  by  substituting  this  value  in  place  of  x  in  2  a;  —  3  and  see  if 
you  get  zero.     In  other  words  see  if  /  (|)  =  0. 

Algebraic  solution:  2^—3  =  0 

2x  =  S, 

x  =  l 

Notice  the  correspondence  of  answers  in  the  geometric  and  the 
algebraic  solutions. 

Example  2.     Let  /  (x)  =2x2  —  3x— 9.      Pqj.  ^hat  value  of  x  is 
/  (x)  =  0.     We  are  to  find  x  so  that  2x2-3x-9  =  0. 


232 


ALGEBRA  — FIRST  COURSE 


tXV,  §  100 


Draw  the  graph  of  the  function  2  x^  —  S  x  —  9.     It  will  be  a 
parabola. 

fix)  X  f{x) 

+  18  -  2  +5 

-  10  +2         -  7 


X 

-3 

+  1 


X 

fix) 

X 

/(x) 

-1 

-4 

0 

-    9 

+  3 

0 

+  4 

+  15 

The  curve  crosses  the  x-axis  where  x  =  —  1.5  and  where  x  =  S. 
These  are  the  required  solutions.     Explain  why. 
Check  by  showing  that  /  (-  1.5)  =0.     Also  that/  (3)  =  0. 
Algebraic  solution:  By  formula,  since  a  =  2,  6  =  —  3,  c  =  —  9, 

3  ±  V9  +  72 


=  3  or  -  |. 


XV,  §  100] 


USES  OF  THE  GRAPH 


233 


Note  the  correspondence  of  the  geometric  and  the  algebraic  solu- 
tions. 

Example  3.     Letf{x)=2x^  +  ix  +  l. 


X 

3 

fix) 

31 

2 

17 

1 

7 

0 

1 

-1 

-  1 

-2 

+  1 

-3 

7 

The  figure  gives  x  =  —  .3 
or  x  =  —  1.7  approxi- 
mately. 

If  we  wish  to  find  al- 
gebraically where  the 
curve  crosses  the  a:-axis, 
solve  for  x  as  in  the  pre- 
ceding exercises.  You  ~l 
should  obtain  the  answers 


=  -1+^V2, 


X 


1  -  i  V2. 


Drawing  these  we  have: 


M  (■ 


O' 


'-.3  or -1,7 


OM    is  the  length  as  expressed  by  —  1  +  ^  V2, 
O'M'  is  the  length  as  expressed  by  —  1  —  ^  V2. 

Comparing  these  with  the  distances  from  the  origin  to  where  the 
curve  in  the  figure  above  crosses  the  x-axis,  we  find  that  they  are  the 
same. 

Check  these  answers  as  you  have  done  in  preceding  exercises. 


234  ALGEBRA  —  FIRST  COURSE  Ixv,  §  lOO 

Exercises.  Give  graphic  and  algebraic  solutions  of  the  fol- 
lowing. By  counting  if  possible,  by  drawing,  if  not,  show 
that  the  geometric  and  algebraic  solutions  agree. 

1.  a;2  -  5 X  +  6  =  0.  9.  Sx^  -  Qx  +  2  =  0. 

2.  rr2  H-  5  x  +  6  =  0.  10.  5  a;^  +  2  a:  -  2  =  0. 
Z.  x^-  x-Q  =  0.  11.  3  x2  -  4  X  -  9  =  0. 
4.  a;2  -  4  =0.  12.  -  x^  +  5 a;  +  1  =  0. 
6.  4  a;2  -  14  =  0.  13.  2x^  -  x  =  0. 

6.  9x2-20  =  0.  14.   7a:2  +  22x  =  0. 

7.  6x2- 7x- 20  =  0.        15.    -3x2-14x  =  0. 

8.  7x2+10x  +  2=0.        16.    -4x2H-20x  =  0. 

17.  Physics  tells  us  that  the  distance  a  body  falls  during 
any  stated  length  of  time  is  expressed  by  the  equation 

where  d  is  in  feet,  t  is  in  seconds,  and  y  =  32  approximately. 

Construct  a  drawing  by  which  you  can  determine  the 
distance  passed  over  by  a  body  faUing  from  rest  during  the 
first  13  seconds;  during  the  first  7  seconds. 

How  long  would  it  take  a  stone  to  fall  from  the  top  of  the 
Washington  monument  to  the  ground? 

Discussion  of  the  quadratic  expression  and  its  graph.  When 
you  made  the  graphic  representation  of  the  quadratic  ex- 
pressions, §  96,  your  attention  was  called  to  the  fact  that  all 
of  the  curves  have  similar  forms,  but  take  different  positions 
with  reference  to  the  axes  and  that  some  are  broader  than 
others. 

As  such  expressions  differ  only  in  the  values  of  a,  b  and  c, 
this  must  be  the  cause  of  the  variations  in  the  position  curve 
and  the  width  of  the  curve. 

We  shall  now  make  drawings  to  find  out  what  effect  the 
changing  of  each  in  turn  has  upon  the  curve. 


XV,  §  100] 


USES  OF  THE  GRAPH  235 


Exercise  1.     Using  the  same  axes,  give  graphic  solution  of  : 
(a)  x2  -  2x  -  2  =  0.         {d)  x^-2x  +  l=0. 


(b)  0:2 -2a;-  1  =0. 

(e)  x2-2x  +  2  =  0. 

(c)   x'-2x          =0. 

(/)  x'-2x  +  S=0. 

You  have  changed  the  value  of  c.  What  effect  does  this 
have  upon  the  curve? 

When  c  is  zero,  through  what  point  of  interest  does  the 
curve  pass ?  Do  you  think  that  the  graph  of/  (x)  =  ax^  +  bx 
will  pass  through  the  origin,  no  matter  what  values  a,  b, 
take  on  ?  Show  this  to  be  true  by  finding  the  roots  of  the 
equation 

ax^  -{-bx  =  0. 

Through  what  other  point  will  the  curve  for /(a;)  =  ax^  +  bx 
always  pass?     Fix  these  truths  well  in  mind. 

Exercise  2.  Make  the  graph  for  the  solution  of  the  follow- 
ing equations  with  reference  to  the  same  axes.  (Not  the 
same  as  used  in  Exercise  1.) 

(a)  x^  -2x-l  =0. 
(6)  x2  -     a:  -  1  =  0. 

(e)  x^-i-2x 

In  this  exercise  you  have  changed  the  value  of  b.  What 
effect  does  this  have  upon  the  curve? 

When  b  is  zero,  what  is  the  fact  of  interest  about  the 
position  of  the  curve  ? 

When  b  and  c  are  both  zero,  what  is  true  about  the  curve  ? 

Write  the  quadratic  equation  with  b  and  c  both  zero,  and 
find  by  algebra  the  value  of  x  for  the  points  where  the  curve 
crosses  the  a:-axis.     How  many  points  are  there? 

Exercise  3.  With  reference  to  the  same  axes  (not  the 
same  as  used  in  Exercises  1  and  2)  make  graphs  for  the 
solution  of  the  following: 

(a)  2x^-Sx~l=0.  (c)     .lx2-3a;-l  =0. 

(b)  x^-dx-l=0.  (d)  .01  x2  -  3a;  -  1  ==  0. 


(c) 

x^-1 

=  0. 

id) 

x^-i-x- 

■1 

=  0. 

1  = 

=  0.   . 

236  ALGEBRA  —  FIRST  COURSE  [XV.§ioi 

In  this  exercise  you  have  changed  a.  What  effect  does 
this  have  upon  the  curve  ? 

What  will  be  true  of  the  graph  if  you  continue  to  make  a 
more  nearly  equal  to  zero  ?  What  would  be  true  if  you  made 
a  zero  ? 

Suppose  a,  b,  c  were  all  zero,  what  would  be  the  graph  of 
the  equation? 

Make  a  summary  of  the  truths  brought  out  in  the  preced- 
ing exercises,  with  reference  to  the  graphs  used  to  solve  the 
following  equations. 

ax^  ■jrbx  =  0. 

ax^  +  c  =  0. 

bx  +  c  =  0. 

ax2  =  0. 

State  these  truths  in  words  and  fix  them  in  mind. 

101.  Nature  of  the  Roots  of  a  Quadratic  Equation.  The 
next  thing  of  interest  about  the  quadratic  equation  is  to  be 
able  to  tell  by  examining  the  equation  whether  or  not  the 
curve  will  cross  the  a;-axis. 

You  will  notice  by  studying  the  curves  of  Exercise  1  that 
some  do  and  some  do  not  cross  the  axis. 

We  shall  now  examine  the  algebraic  equation  and  its  roots 
and  determine  the  cause  of  this. 

The  roots  of 

ax^  +  bx  -\-  c  =  0 
are  x  = 


-  6  +  V52  - 

-  4ac 

2a 

-b-Vb^- 

-  4ac 

^~  2a 


Examining  these  roots  we  see  that  the  only  difference  in 
them  is  the  sign  of  Vb^  —  4  ac.    If  this  quantity  be  added 


XV.  §1011  QUADRATIC  EQUATIONS  237 

to  —  6  and  the  sum  divided  by  2  a,  we  get  one  root,  or  one 
crossing  point  of  the  curve  with  the  a:-axis.  If  we  subtract 
this  quantity  from  —  b  and  divide  by  2  a  we  get  the  other 
root,  or  the  other  crossing  point. 

Now  b^  —  4  ac  will  be  positive,  zero,  or  negative,  accord- 
ing to  the  relative  values  of  a,  b,  c. 

If  b^  —  4:ac  is  greater  than  zero,  the  two  roots  are  unequal 
and  the  curve  crosses  the  a:-axis  in  two  distinct  points, 
whose  distances  from  the  origin  can  be  counted,  or  can  be 
drawTi  by  making  use  of  the  Pythagorean  theorem.  Ex- 
amples 2  and  3,  of  §  100,  illustrate  this.  In  Example  2  the 
roots  are  rational;  in  Example  3  they  are  irrational. 

See  Exercise  1  (a),  (6),  (c).  Examine  the  graphs  and  com- 
pute the  value  of  6^  —  4  ac  to  see  if  the  above  statements 
hold. 

7/6^  —  4  ac  is  zero,  the  roots  will  be  equal,  since  adding 
zero  to  —  6  gives  the  same  result  as  subtracting  zero  from 
—  b.     In  this  case  the  parabola  is  tangent  to  the  a:-axis. 

Examine  Exercise  1  (d),  to  see  if  the  above  statements 
hold.  

7/62  —  4  ac  is  negative,  Vb^  —  4:ac  is  imaginary;  this 
means  that  the  roots  are  imaginary,  and  the  parabola  shows 
no  point  in  common  with  the  x-axis. 

See  Exercise  1  (e)  and  (/).  Examine  as  to  the  truth  of 
the  preceding  statements. 

To  sum  up: 

when  &^  —  4  ac  is  positive,  the  roots  are  real  and  unequal; 
when  6^  —  4  ac  is  zero,  the  roots  are  real  and  equal; 
when  6^  —  4  ac  is  negative,  the  roots  are  imaginary. 

Fix  in  mind  the  above  facts  about  the  roots  of  the  quad- 
ratic equation  and  the  position  of  the  graph  with  reference 
to  the  X-axis. 

Your  attention  is  called  to  another  feature  of  interest. 


238  ALGEBRA  —  FIRST  COURSE  lxv,§io2 

You  will  notice  that  if  you  should  draw  a  line  parallel  to  the 
function  axis,  through  the  vertex  of  your  curve,  it  will  divide 
it  in  such  a  way  that  if  you  revolve  the  curve  through  a 
straight  angle  about  this  line  as  an  axis,  the  curve  will 
fall  in  identically  the  same  place  that  it  was  before  you 
revolved  it. 

The  line  is  called  the  axis  of  symmetry,  and  the  curve  is 
said  to  be  symmetrical  to  this  line.  The  equation  of  this 
line  is 

_  -h 
^~  2a' 

By  drawing  in  this  line  you  can  test  to  see  that  you  have 
your  curve  well  drawn,  for  you  should  be  able  to  start  with 
any  point  on  your  curve,  draw  a  perpendicular  to  the  axis 
of  symmetry,  continue  in  the  same  direction  for  the  same 
distance  and  strike  another  point  on  your  curve.  In  other 
words  all  lines  which  cross  your  curve  perpendicular  to  the 
axis  of  symmetry  must  be  bisected  by  it.  Otherwise  your 
curve  is  not  properly  drawn. 

Another  point  of  less  interest,  but  which  will  aid  in  your 
drawing,  is  that  the  curve  crosses  the  function  axis  at  the 
point  c  units  from  zero.     Why? 

Examine  the  equations  on  p.  234,  and  tell  all  you  can 
about  the  position  of  the  graph  with  reference  to  the  axes, 
the  nature  of  the  roots,  and  the  axis  of  symmetry. 

102.   Graphic  Solution  of  Equations  of  Higher  Degree. 

Example:  Let/ (a;)  =  x^  —  Z  x^  —  3  x  -\-  9.    For  what  values  of  x 
does  /  (x)  equal  zero  ? 
We  are  to  find  x  so  that 

a:3  -  3  a;2  -  3  X  +  9  =  0. 

The  algebraic  solution  of  such  an  equation  is  rather  complicated, 
unless  we  can  factor /  (x) .  Can  you  do  this?  We  pass  to  the  graphic 
solution. 


XV,  §  1031 


SUMMARY 

X 

/(x) 

X 

/(*) 

-2 

-    5 

2 

-    1 

-1 

+    8 

3 

0 

0 

+    9 

4 

+  11 

1 

+    4 

239 


By  inspection  of  the  graph  we  see  that  f  {x)  =0  when  x  =—  1.7 
approximately,  when  x  =  +1.7  approximately,  and  when  x  =  3.  By 
greatly  enlarging  the  part  of  the  graph  near  x  =  1.7,  say  from  x  =  1.5 
to  X  =  2,  can  you  get  the  second  decimal  place  in  this  value  of  x?  Do 
this. 

Exercises.  Obtain  graphic  solutions  of  the  following 
equations.     Give  exact  solutions  when  possible,  by  factoring. 


1.  x^  -4:X  =  0. 

2.  x^  -2x^  +  x  =  0. 

3.  4a;3  -  9  a;^  =  0. 


•3  _ 


x^  +  x-1  =0. 


a;3  +  a;2  -  9  X  -  9  =  0. 


x^  -\-  x^  —  9x 
a;*  -  5  a;2  +  4 
x^  =  4  X  +  4. 


5=0. 
0. 


103.   Summary. 

The  practical  use  of  the  graph  is  to  represent  tabulated 
values  of  a  function. 


240  ALGEBRA  —  FIRST  COURSE  [XV,  §103 

Graphs  are  also  used  to  solve  equations.  Draw  the  graph 
of  the  function  and  measure  the  values  of  the  variable  to 
the  points  where  the  graph  crosses  the  axis  of  the  variable. 
These  values  are  the  solutions.  When  there  are  no  crossing 
points  the  solutions  are  imaginary. 

To  solve  graphically  the  linear  equation  ax  -{-  h  =  0,  draw 
the  graph  of  ax  +  6  and  note  the  point  where  the  straight 
line  crosses  the  x-axis. 

To  solve  the  quadratic  equation  ax^  -\- hx -\-  c  =  0  graphi- 
cally, draw  the  graph  of  the  expression  ax^  +  hx  -\-  c  and 
note  the  points  where  it  crosses  the  x-Sixis. 

This  graph  is  a  parabola  whose  position  depends  on  the 
numerical  values  of  a,  b  and  c. 

If  ?>2  —  4  dc  is  positive,  the  parabola  cuts  the  x-axis  twice. 
In  this  case  the  quadratic  has  two  real  and  unequal  solutions. 

If  6^  —  4  ac  is  zero,  the  parabola  just  touches  the  a;-axis. 
In  this  case  the  quadratic  has  two  equal  solutions. 

If  62  _  4  ac  is  negative,  the  parabola  does  not  cross  the 
a;-axis.     In  this  case  the  roots  are  imaginary. 


CHAPTER  XVI 

LOCI    OF   POINTS.     SIMULTANEOUS   EQUATIONS* 

104.  Meaning  of  the  Word  Locus.  You  cannot  locate  a 
place  without  mentioning  its  relative  position  with  reference 
to  some  other  place.  And,  furthermore,  as  the  following 
examples  will  illustrate,  you  cannot  locate  a  place  without 
giving  two  descriptions  of  its  location. 

Example  1.  I  wish  to  tell  you  the  location  of  a  house.  I  say  that 
it  is  a  mile  from  here.  Immediately  as  you  think  of  houses  one  nule 
from  here  they  arrange  themselves  on  the  circumferemce  of  a  circle 
which  has  our  present  location  for  a  center  and  one  mile  for  a  radius. 
Any  house  nearer  or  farther  as  you  are  able  to  estimate  it,  is  put  out 
of  mind.  There  may  be  many  houses  which  answer  the  description 
but  they  all  stand  on  the  circumference.  The  circumference  is  the 
place  of  the  house.  Mathematically  I  would  say  that  the  circumfer- 
ence is  the  locus  of  the  house  described  as  being  one  mile  from  here. 
The  word  locv^  means  place.  As  is  readily  seen,  one  description  is 
not  sufficient  to  locate  the  house  that  I  have  in  mind.  I  will  say  further 
that  the  house  is  three  blocks  from  the  street  that  runs  in  front  of  the 
school  house.  Immediately  your  .mind  runs  along  a  street  three 
blocks  away  and  when  it  reaches  what  you  estimate  to  be  a  mile  from 
here  you  have  in  mind  the  house  we  are  speaking  of.  The  street  is  a 
second  place  or  locus  of  the  house  and  where  the  two  loci  come  to- 
gether is  the  place  where  the  house  stands.  There  might  be  two  such 
houses.     Explain  how  so. 

Example  2.  Locate  a  place  when  the  description  is  given  as  105° 
west  longitude,  and  41°  north  latitude. 

As  you  learned  in  your  geography,  for  the  sake  of  locating  places  on 
the  earth's  surface,  a  line  of  reference  is  supposed  to  be  drawn  through 
Greenwich,  running  north  and  south  from  pole  to  pole.  Also  a  second 
line  of  reference,  the  equator,  runs  at  right  angles  to  this.     A  place 

*  For  closer  correlation  this  chapter  may  be  preceded  by  Chapter  V 
of  Geometry. 

241 


242 


ALGEBRA  —  FIRST  COURSE 


(XVI,  §  105 


whose  description  is  given  as  105°  west  longitude  might  be  anywhere 
along  the  meridian  105°  west  from  the  line  of  reference.  If  we  use  a 
mathematical  term,  the  meridian  would  be  its  locus.  A  place  de- 
scribed as  41°  north  latitude  might  be  anywhere  along  the  parallel  41° 
north  of  the  equator.  This  parallel  would  be  its  locus.  Where  these 
two  lines  cross  would  be  the  place  sought. 

Give  other  examples  of  ways  for  locating  places. 


105.  Coordinates.  To  locate  the  position  of  a  point  on 
a  sheet  of  paper  or  any  plane  surface,  we  follow  the  same 
plan  as  is  used  to  locate  a  place  on  the  earth's  surface.  Draw 
two  lines  of  reference  which  cut  at  right  angles;  one  extend- 
ing from  left  to  right  and  the  other  up  and  down.  (Always 
use  cross-section  paper  for  this  work.) 

The  plane  of  the  paper  is  now  divided  into  four  quarters, 
called  quadrants.     They  are  numbered  I,   II,   III,   IV  as 

shown  in  the  figure. 
We  now  locate  a  point 
by  giving  its  distance 
from  the  two  reference 
lines,  stating  also 
whether  these  distances 
are  measured  to  the 
right  or  left,  upward 
or  downward.  The  line 
x'x,  as  shown  in  the 
figure,  is  called  the  axis 
of  the  abscissa.  The 
line  y'y  is  called  the 
axis  of  the  ordinate. 
Distances  upward  from 
the  axis  of  the  abscissa  are  positive;  downward,  negative. 
Distances  to  the  right  of  the  axis  of  the  ordinate  are  positive; 
to  the  left,  negative.  The  point  of  intersection  marked  zero 
is  called  the  origin.  The  distance  which  a  point  is  to  the 
right  or  left  of  the  axis  of  the  ordinate,  measured  in  the 


y 

■'■ 

1 

J 

T 

r^ 

._ 

■" 

- 

f 

, 

"o 

1 

^r 

'  J 

~1 

J 

11 

1 

I  \ 

t 

± 

— 



i-W4 

-i 

I 

-- 

- 

- 

y 


XVI.  §  105] 


LOCI 


243 


direction  of  the  axis  of  the  abscissa,  is  called  the  abscissa  of 
the  point.  The  distance  which  a  point  is  from  the  axis  of 
the  abscissa,  measured  in  the  direction  of  the  axis  of  the 
ordinate,  is  called  the  ordinate  of  the  point.  The  two  to- 
gether are  called  the  coordinates  of  the  point  which  is  fixed 
by  them. 

We  usually  represent  the  abscissa  of  a  point  by  x,  and  its 
ordinate  by  y;  then  the  two  numbers  x,  y  are  the  coordinates 
of  the  point.     However,  any  other  letters  may  be  used. 


Example.     Suppose  we  wish  to  locate  a  point  for  which  x  =  2  and 
2/  =  5,  that  is,  the  abscissa  is  2  and  the  ordinate  is  5. 
Since  the  abscissa  is  2, 


y= 


2/- 


(^,5) 


^ 


the  point  lies  2  units  to 
the  right  of  the  axis  of 
the  ordinate.  It  is  there- 
fore on  a  line  drawn  2  - 
units  to  the  right  of  the 
axis  of  the  ordinate  and  ~ 
parallel  to  that  axis.  This 
line  we  designate  by  the 
equation  x  =  2,  because 
this  equation  is  true  for 
any  point  on  the  line,  and 
is  true  for  no  other  point. 
This  line  is  the  locus  of 
the  point  whose  abscissa 
is  2. 

Since  the  ordinate  of 
the  point  is  5,  the  point 
must  be  somewhere  on  a 

line  drawn  parallel  to  the  axis  of  the  abscissa  and  5  units  above  it. 
This  hne  is  designated  by  the  equation  y  =  5.  Why?  This  hne  is 
the  locus  of  a  point  whose  ordinate  is  5. 

The  required  point  must  be  at  the  intersection  of  the  loci  just  drawn. 
We  call  it  the  point  (2,  5);  here  we  inclose  in  parentheses  the  coordi- 
nate of  the  point,  writing  first  the  abscissa,  and  then  the  ordinate,  with 
a  comma  between. 

A  point  whose  abscissa  is  z  and  whose  ordinate  is  j/  is  designated  by 
the  symbol  (x,  y). 


244  ALGEBRA  —  FIRST  COURSE  [xvi,  §  106 

Exercises.  Draw  the  lines  on  which  each  of  the  following 
points  must  lie.  On  each  line  write  the  equation  which 
describes  it.  As  has  just  been  explained  such  a  line  is  the 
locus  of  the  point  which  lies  on  it. 

1.  What  is  the  locus  of  a  point  whose  abscissa  is  3? 
Whose  ordinate  is  7?  What  point  is  at  the  intersection  of 
the  two  loci  ?  Draw  a  complete  figure  as  shown  above,  mark- 
ing in  the  equations  of  the  line  and  the  symbol  for  their  point 
of  intersection. 

2.  Proceed  as  in  Exercise  1,  when  the  abscissa  is  3  and  the 
ordinate  is  —  7. 

3.  Proceed  as  in  Exercise  1,  when  the  abscissa  is  —  5  and 
the  ordinate  is  2. 

4.  Proceed  as  in  Exercise  1,  when  the  abscissa  is  —  5  and 
the  ordinate  is  —  2. 

5.  Proceed  as  in  Exercise  1,  when  the  abscissa  is  —  3  and 
the  ordinate  is  —  3. 

6.  Make  diagrams  as  in  Exercise  1  for  each  of  the  follow^- 
ing  points  : 

(2,  6);   (-  5,  4);   (6,  -  9);  (12,  -  7);   (-  2,  -  3);  (0,  -  9); 
(2,  0). 

7.  What  line  is  described  by  the  equation  x  =  0?  What 
line  is  described  by  the  equation  ?/  =  0?  What  point  lies 
at  the  intersection  of  these  loci  ? 

8.  Locate  the  points  (9,  4);  (—  3,  —  1);  (4,  —  3).  Join 
by  straight  lines.     What  kind  of  a  figure  is  formed  ? 

9.  Draw  the  quadrilateral  whose  vertices  are  the  points 
(3,  4),  (-  1,  4),  (-  1,  -  2),  (3,  -  2).  How  long  is  the 
diagonal  of  this  quadrilateral  ? 

106.  Straight  Line  Loci  in  General.  We  shall  study  ex- 
amples of  loci  which  are  not  parallel  to  one  of  the  reference 
lines  in  the  exercises  that  follow.  Use  the  same  sheet  of 
paper  for  Exercises  1  to  7  inclusive,  drawing  the  axis  rather 
heavy  near  the  middle  of  the  sheet  each  way. 


XVI.  §  1061  LOCI  245 

Exercises. 

1.  Bisect  the  pair  of  vertical  angles  made  by  the  axes  of 
the  coordinates  which  form  the  first  and  third  quadrants. 
Place  five  points  along  this  bisector,  some  on  one  side  of  the 
origin  and  some  on  the  other.  Count  the  abscissa  and  the 
ordinate  of  each  point.     How  do  they  compare  in  each  case? 

The  algebraic  statement  of  this  thought  is 

X  =  y,  or  X  —  y  =  0, 
Write  the  equation  on  the  line. 

2.  Draw  the  line  y  =  I.  Bisect  the  pair  of  vertical  angles 
formed  by  the  lines  y  =  1  and  x  =  0,  corresponding  to  the 
pair  bisected  in  Exercise  1.  Place  5  points  along  this  line, 
some  on  one  side  of  the  point  of  intersection  and  some  on  the 
other.  Ascertain  by  counting  the  comparative  lengths  of 
the  abscissa  and  ordinate  of  each  point.  (Remember  to 
count  from  the  axes  of  the  abscissa  and  ordinate  each  time.) 
Explain  why  you  arrive  at  the  result  that  you  do  each  time. 

The  algebraic  statement  of  this  thought  is 
y  =  x-{-l. 

3.  Repeat  the  work  of  Exercise  2,  bisecting  the  angle 
formed  by  the  lines  y  =  2,  x  =  0;  y  =  3,  x  =  0;  y  =—  1, 
X  =  0;  y  =  —2,  x  =  0.  On  each  bisector  write  the  algebraic 
expression  of  the  truth,  found  by  counting,  concerning  the 
relation  of  the  coordinates  of  each  point. 

4.  These  bisectors  look  parallel.  We  shall  assume  that 
they  are  parallel. 

5.  Bisect  the  other  pair  of  vertical  angles  formed  by  the 
lines  y  =  0,  X  =  0.  By  counting  estabhsh  the  relation  of 
the  abscissa  and  ordinate  of  points  on  this  line. 

The  algebraic  expression  for  this  thought  is 

y  =—  X,  or  y  -\-  X  =  0. 
Write  this  name  on  the  line. 

6.  Draw  the  bisectors  of  the  other  pairs  of  vertical  angles, 
and  by  counting  establish  the  relation  of  abscissa  and  ordi- 


246  ALGEBRA  —  FIRST  COURSE  Ixvi.  §  106 

nate  of  points  on  each,  express  in  algebraic  language,  and 
write  the  name  on  each  line. 

7.  To  generalize  the  idea  brought  out  in  Exercises  1  to  6, 
consider  y  =  b  to  he  the  line  parallel  to  the  axis  of  the 
abscissa  (6  being  any  distance  up  or  down  from  the  axis  of 
the  abscissa).  The  bisectors  of  the  angles  formed  by  y  =  b 
and  X  =  0,  will  be  parallel  to  the  bisectors  of  the  angles 
formed  by  y  =  0  and  x  =  0.  This  being  true,  we  can  state 
that  the  general  name  of  these  lines  is  y  =  x  -\-  b  for  the 
first  set,  and  y  =  —  x  -{•  b  for  the  second.  On  a  new  page 
of  graph  paper,  with  axes  of  reference  drawn  as  before, 
place  Exercises  8,  9  and  10. 

8.  Starting  at  the  origin  to  count,  mark  a  point  whose 
abscissa  is  1  and  whose  ordinate  is  2.  Join  this  point  to  the 
origin,  extending  the  line  indefinitely  in  either  direction. 
By  counting  establish  as  in  previous  cases  the  relation  of  the 
abscissa  and  ordinate  of  points  on  this  line.  You  find  in 
each  case  the  ordinate  to  be  2  times  the  abscissa. 

The  algebraic  expression  for  this  relation  is 

y  =2x. 

9.  Starting  from  the  point  of  intersection  oi  y  =  1  and 
X  =  0,  count  1  unit  to  the  right  and  2  units  up.  Join  the 
point  to  the  point  of  intersection  oi  y  =  1  and  x  =  0,  ex- 
tending the  line  indefinitely. 

By  counting  establish  the  relation  of  abscissa  and  ordinate 
of  points  on  this  line. 
The  algebraic  expression  of  this  relation  is 

y  =  2x  +  h 

In  every  case  be  sure  to  write  names  on  the  lines. 

10.  Repeat  instructions  of  Exercise  9,  counting  from  the 
point  of  intersection  of  the  following. 

y  =  2  and  x  =  0;  y  =  S  and  x  =  0;  y  =—  1  and  x  =  0; 
2/  =  —  2  and  x  =  0. 


XVI.  §  106J  LOCI  247 

By  a  discussion  as  in  Exercise  7,  bring  out  the  statement 
that  the  general  algebraic  expression  for  the  equation  of 
these  lines  is 

y  =  2x  +  h, 

where  h  is  any  distance  measured  up  or  down  from  the 
origin. 

On  a  new  page  of  graph  paper  place  Exercises  11,  12,  13. 

11.  Repeat  the  instructions  of  Exercises  8,  9,  10,  estab- 
lishing the  line  by  counting  1  to  the  right  and  3  up  from  the 
origin  and  the  points  of  intersection.  You  should  be  able 
to  write  the  names  on  these  lines  without  any  trouble.  Do 
this  work  on  a  fresh  sheet  of  paper. 

Write  the  algebraic  expression  for  the  relation  of  the 
abscissa  and  ordinate  of  the  points  on  each  line. 

12.  Repeat  Exercises  8  to  11  counting  to  the  left  and  up, 
instead  of  to  the  right.  Write  the  names  on  these  lines 
without  counting.  Then  count  for  one  point  to  see  that  you 
have  written  correctly. 

13.  We  shall  now  try  to  write  an  algebraic  equation, 
which  will  be  a  general  expression  of  the  idea  brought  out  by 
Exercises  1  to  12.  Examine  your  equations  and  you  will  see 
that  the  coefficient  of  x  is  the  expression  of  the  ratio  oi  y  to  x 
on  the  line  through  the  origin.  That  is,  it  gives  you  the  slope 
of  the  line.  All  other  lines  which  have  the  same  coefficient  of 
X  are  parallel  to  it  and  hence  have  the  same  slope.  If  you 
count  1  unit  to  the  right  and  a  units  up,  you  have 

y  =  ax,  a  being  positive  in  this  case. 

If  you  count  1  unit  to  the  left  and  a  units  up,  you  have 

y  —  ax,  a  being  negative  in  this  case. 

The  line  parallel  to  any  one  of  these  lines,  and  passing 
through  the  intersection  oi  y  =  b  and  x  =  0,  has  for  its 
equation 

y  =  ax-\-b. 


248  ALGEBRA  —  FIRST  COURSE  [xvi.  §  106 

This  equation  is  very  general.  You  cannot  draw  any 
straight  Une  on  your  paper  the  relation  of  whose  coordi- 
nates this  equation  will  not  describe.  Moreover  you  cannot 
locate  a  point  on  your  paper  that  will  not  fall  on  one  or  more 
of  these  lines;  that  is,  you  cannot  locate  a  point  whose  co- 
ordinate will  not  satisfy  this  equation. 

Being  given  a  line,  we  have  learned  how  to  write  its 
equation.  We  shall  now  investigate  the  converse  of 
this. 

Being  given  an  algebraic  expression,  to  determine  the  line 
which  is  its  graph.  We  shall  do  this  by  examining  a  special 
case. 

Example  1.     Given  the  equation  2x  -\-  Sy  =  0. 
To  find  the  locus  of  points  whose  coordinates  satisfy  this  equation. 
The  equation  may  be  written  in  the  form 


Comparing  this  with  the  equation  y  =  ax,  a  =  —  f ;  so  we  can 
locate  one  point  on  the  line  by  counting  3  to  the  left  and  2  upward, 
that  is,  locate  the  point  (—3,  2).  Draw  a  straight  hne  through  this 
point  and  the  origin,  extending  it  each  way  indefinitely,  and  we  have 
the  graph  of  the  equation  2  x  +  3  ?/  =  0. 

Test  this  by  selecting  any  other  point  on  this  line,  counting  its 
coordinate  distances,  and  see  if  they  satisfy  the  equation  given. 

Example  2.     Given  the  equation  2  x  +  3  ?/  =  19. 

Subtracting  2x  from  both  members  of  the  equation,  and  dividing 
by  3,  this  equation  becomes, 

y  =-fa;  +  6i 

We  now  have  i/  as  a  function  of  x,  of  the  form  ax  +  h. 
Here  we  have  a  =  —  f ,  and  6  =  65. 

To  make  the  graph  locate  the  point  where  the  line  y  =  Q^  intersects 
the  line  x  =  0.  From  this  point  count  3  units  to  the  left  and  2  up- 
ward. Join  these  two  points  and  this  line  extended  indefinitely  in 
either  direction  is  the  graph  of  the  given  equation.  Test  this  by 
selecting  another  point  on  the  line  and  counting  to  determine  its  co- 
ordinates.    See  if  they  satisfy  the  equation. 

Compare  this  work  with  the  work  in  §  95. 


XVI,  §  106] 


LOCI 


249 


) 

N 

^^ 

s 

N 

N 

y-6 

/s 

N. 

s 

V 

^^ 

s. 

s 

\ 

\ 

(-^: 

2) 

\ 

^ 

S^ 

% 

\ 

0 

V 

s^ 

"!» 

N 

V 

Exercises.     By  the  process  shown  above  draw  the  locus  for 
each  of  the  following  relations  between  x  and  y. 

1.  X  —  y  =  0. 

2.  X  -\-  y  =  1. 

3.  6x  +  Sy  =  0. 

4.  Qx-\-Sy  =  7. 
6.  4  a;  +  3 1/  =  0. 

6.  4  a;  -f  3  2/  =  12. 

7.  3  X  -  4  2/  =  0. 

These  equations  are  all  of  the  form 
ax  -^  hy  -\-  c  =  0, 

where  the  letters  a,  h  and  c  stand  for  given  numbers,  positive 
or  negative.  Such  an  equation  is  called  an  equation  of  the 
first  degree  or  a  linear  equation  in  two  variables. 


8. 

3x-4:y  =-  14. 

9. 

2x-y  =  3. 

10. 

x-2y  =  3. 

11. 

x-22/  +  4  =  0. 

12. 

Sx-2y  =  Q. 

13. 

2x  +  3y  +  Q  =  0. 

14. 

x-y  =  2. 

250  ALGEBRA  —  FIRST  COURSE  [xvi,  §  107 

From  a  study  of  the  preceding  exercises,  what  kind  of  a 
locus  is  determined  by  a  linear  equation? 

Does  your  work  suggest  the  following  conclusions  ? 

(a)  Any  point  whose  coordinates  x  and  y  satisfy  the  given 
equation  lies  on  the  locus  of  that  equation. 

(6)  Conversely,  if  we  read  off  the  coordinates  of  any  point 
on  the  locus,  we  shall  have  a  pair  of  values  of  x  and  y  for 
which  the  equation  is  true. 

In  other  words,  the  locus  of  the  equation  gives  us  a  com- 
plete geometric  picture  of  all  the  pairs  of  values  of  x  and  y 
which  make  the  equation  true. 

107.  Simultaneous  Equations.  Since,  as  we  have  seen, 
one  equation  gives  us  a  locus  of  points  whose  coordinates 
satisfy  the  equation,  that  is,  one  equation  gives  us  one  rela- 
tion between  x  and  y,  if  we  wish  to  locate  a  definite  point,  it 
will  be  necessary  to  have  two  equations,  so  that  we  may 
have  two  relations.  The  coordinates  of  the  point  or  points 
of  intersection  of  the  loci  will  then  satisfy  both  relations. 

Definition,  Two  equations  which  are  satisfied  by  the 
same  pair  of  values  of  x  and  y  are  called  simultaneous  equa- 
tions. 

Example.     Given  the  two  equations 

(1)  2  a:  +  3  ?/  =  19, 

(2)  3  a;  -  4  ?/  =  -  14. 

To  find  by  graph  the  point  whose  coordinates  satisfy  the  two  equa- 
tions. 

Graphic  Solution:  Solve  each  for  y  in  terms  of  x,  or  writing  y  =  f  (x). 

(1)  y  =  -ix  +  Qh 

(2)  y  =  ix  +  3h 

Drawing  the  graph  of  these  as  shown  in  the  preceding  article,  we 
have  the  figure  on  the  next  page. 

Check  this  by  substituting  the  values  2  and  5  for  x  and  y  respec- 
tively in  the  given  equations  to  see  if  they  verify  them. 

Thus   2'2  +  3'5  =  4  +  15  =  19,  as  the  first  equation  states. 

3'2  —  4'5  =  6  —  20=— 14,  asthe  second  equation  states. 


XVI,  §  107] 


SIMULTANEOUS  EQUATIONS 


251 


t 

N 

kj. 

s; 

% 

p 

y 

y 

\^ 

s 

y--  6h 

J 

Y 

Nv 

N 

N 

\y 

'2,5) 

> 

y 

M 

s. 

ly 

N 

A 

y 

-s'A 

N 

<-\ 

y 

^ 

k 

y 

/^" 

W4 

y 

0 

^X 

Algebraic  solutions:  We  may  obtain  the  coordinates  of  the  point  of 
intersection  by  algebra  as  well  as  by  drawing. 

(a)  Solution  by  Comparison.  Observe  that  for  the  point  of  inter- 
section and  for  no  other  point  on  the  lines,  the  value  of  y  is  the  same 
for  the  same  value  of  x,  whether  we  regard  the  point  as  belonging  to 
one  locus  or  the  other. 

Therefore,  since  y  =  —  f  a;  +  6^, 

and  also  2/  =  f  ^  +  3^, 

it  follows  that  for  the  point  of  intersection  and  for  this  point  only 

Multiplying  both  members  by  12,  we  have 

-8a;  +76  =  9x  +  42, 
17  X  =  34, 
x  =  2. 

Substituting  this  value  for  x  in  the  equation 

2/=  -fx  +  ei 


252  ALGEBRA  —  FIRST  COURSE  [XVI.  J107 

we  have  y  =  —  f  •  2  +  6^ 

=  5. 

These  are  the  coordinates  of  the  point  of  intersection  as  shown  in 
the  drawing.  Check  by  substituting  these  values  in  the  second  equa- 
tion. 

Exercises.  In  each  of  the  following  exercises  make  graph; 
find  the  point  common  to  the  two  loci;  solve  by  algebra  to 
find  the  values  of  x  and  y  that  will  satisfy  both  equations; 
check  to  make  sure  that  your  algebraic  solution  is  correct. 
Compare  these  values  with  the  coordinates  of  the  point  of 
intersection,  and  make  sure  they  are  the  same. 

1.  a;  -  2  2/  =  6,  4.   3  ?/  -  8  a;  =  19, 
7x  +  52/  =  23.  ^x-y      =0. 

2.  4a; +  2/     =  -13,  5.    \x  +  \y  =  12, 
8a;  +  32/  =  -27.  x-y     =4. 

3.  5x  +  42/  =  +2,  6.   fa;  +  t2/  =  21, 
3x  +  2/     =-3.  Ix  +  iy  =  l7. 

In  the  algebraic  solution  when  you  find  the  values  of  y  in 
terms  of  x  and  the  constant,  and  equate  these  values,  you 
eliminate  one  of  the  variables,  namely  y,  and  get  an  equation 
containing  only  one  variable. 

This  special  way  of  eliminating  one  of  the  variables  is 
called  elimination  hy  comparison. 

The  main  idea  in  the  process  is  to  obtain  an  equation 
containing  only  one  variable.  Any  process  that  will  accom- 
plish this  result,  and  not  violate  any  of  the  laws  of  mathe- 
matics, will  answer  our  purpose. 

There  are  two  other  methods  that  may  be  used,  which  we 
shall  now  investigate. 

(6)  Solution  by  Substitution.  Write  one  of  the  two  given  equations 
in  the  form  of  y  =  f  (x),  as  in  the  preceding  method.  Since  for  the 
point  of  intersection,  the  value  of  y  is  the  same  for  each  equation, 
substitute  this  value  of  y  in  place  of  y  in  the  other  equation,  thus 
obtaining  an  equation  containing  one  variable. 


XVI.  §  107]  SIMULTANEOUS  EQUATIONS  253 

Example.     Using  the  same  equations  that  we  have  been  discussing, 
2x-f-3  2/  =  19, 
3x  -  4y  =  -14. 

Writing  the  first  equation  in  the  form  of  i/  =  f  (x), 

y  =  -Ix  +  Qi 

Substituting  this  value  in  the  second  equation  we  have 
dx  -  4(-|a;  +  6i)  =  -14. 
From  which  V-x  =  11^, 

and  X  =  2. 

Since  y  =  —  I  x  +  6^ 

and  X  =  2, 

2/=  -f.2  +  6| 
=  5. 

(2,  5)  is  the  point  of  intersection  as  shown  in  the  graph. 
(c)  Elimination  by  Multipliers.  The  third  method  for  obtaining  an 
equation  containing  only  one  variable  is  the  most  commonly  used.  It 
consists  in  first  multiplying  the  equations  by  numbers  that  will  make  the 
coefficients  of  one  of  the  variables  the  same  in  the  two  equations,  then 
if  the  terms  which  have  like  coefficients  have  like  signs,  we  will  elimi- 
nate one  of  the  variables  by  subtracting  one  equation  from  the  other, 
thus  obtaining  an  equation  of  only  one  variable.  If  the  terms  with 
like  coefficients  have  unlike  signs,  we  can  accomplish  the  same  result 
by  adding  the  two  equations.  What  is  the  axiom  that  is  appUed  in 
either  case? 

This  process  is  often  called  elimination  by  addition  or  subtraction. 
To  illustrate  this  we  shall  use  the  same  equations  which  we  solved 
by  the  other  two  methods, 

2x-f-32/ =  19, 
Sx  -  4:y  =  -14. 
Multiplying  the  first  equation  by  3  and  the  second  by  2,  we  have 
6a; +  9?/ =  57,  Why? 

Qx-Sy  =  -28.  Why? 

Subtracting  one  equation  from  the  other,  we  have, 
17  2/ =  85, 
2/ =  5. 
We  might  have  multiplied  the  first  equation  by  4  and  the  second 
by  3,  thus  obtaining 

8x-i-12y  =  76,  Why? 

9x  -  12y  =  -42.  Why? 


254  ALGEBRA  —  FIRST  COURSE  [xvi.  §107 

Adding  one  equation  to  the  other  (since  the  sign  of  12  y  is  positive  in 
one  equation  and  negative  in  the  other) 

17  a;  =  34.  Why? 

x  =  2. 

The  check  is  the  same  in  all  three  cases. 

This  method  can  best  be  carried  out  by  the  following  scheme. 


3 
-2 


2x  +  Zy  =  19 
Zx  -  Ay  =  -14 


17?/ =  85,     2/ =  5, 
17  X  =  34,     X  =  2. 

Exercises.     Solve  the  exercises  on  p.  252  by  the  last  method. 
Also  solve  the  following: 

1.  aix  +  hiy  =  ci,  Z.    {c-h)x-  y  =  c, 

a^x  +  62?/  =  C2.  h'^x  -\-  (c -\- h)  y  =  —  cb. 

2.  ax  —  y     —2m,  4.  x  +  ay  =  b, 

X  —  by  =  n.  dx  —  y  =  c. 

There  are  sets  of  simultaneous  equations  which  are  not 
linear,  but  which  may  be  solved  as  such. 

In  the  following  regard  -  and  -  as  the  unknowns;  do  not 
X         y 

clear  fractions,  but  multiply  each  equation  by  a  number  that 

will  make  the  coefficients  of  one  of  the  variables  the  same, 

and  add  or  subtract  as  the  case  may  require,  to  eliminate 

one  of  the  variables.     Then  clear  fractions  and  solve  for 

the  other  variable.     By  the  same  process  find  the  value  of 

the  other  variable  and  check  as  in  the  former  work.         • 


6. 


6. 


U5  =  19, 

7. 

11      5 
x^y-Q' 

§-5  =  7. 
X      y 

1      2          5 

X      y           6 

5  +  §  =  3, 

X      y 

8. 

1       1 

1^-^  =  4 
X      y 

?-5  =  6. 

X     y 

XVI.  §  108]  SUMMARY  —  PROBLEMS  255 

108.   Summary  and  Problems. 

The  locus  of  a  point  is  a  line  or  lines  on  which  a  point 
must  fall  in  order  to  fulfil  a  condition  imposed  upon  it. 
Every  point  on  the  locus  fulfils  the  condition  and  every  point 
that  fulfils  the  condition  is  on  the  locus. 

A  plane  is  divided  into  quadrants  by  means  of  two  straight 
lines  drawn  perpendicular  to  each  other.  These  lines  are 
axes  of  reference;  the  one  running  from  left  to  right  is  called 
the  axis  of  the  abscissa;  the  one  running  up  and  down  is 
called  the  axis  of  the  ordinate.  The  point  of  intersection  is 
called  the  origin. 

Every  point  in  the  plane  may  be  located  with  reference 
to  these  straight  lines;  its  distance  from  the  axis  of  the 
ordinate  is  called  the  abscissa  of  the  point;  its  distance 
from  the  axis  of  the  abscissa  is  called  the  ordinate  of  the 
point.  The  two  distances  are  called  the  coordinates  of  the 
point. 

If  a  relation  is  established  between  the  coordinates  of  a 
point,  and  if  this  relation  is  expressed  in  algebraic  symbols, 
the  expression  is  called  an  equation  in  two  variables.  If  the 
graph  is  a  straight  line,  its  equation  is  called  the  linear  equa- 
tion in  two  variables. 

There  are  indefinitely  many  solutions  to  a  linear  equation 
in  two  variables.  The  graphs  of  two  linear  equations  in  two 
variables  are  necessary  to  locate  a  point. 

Two  linear  equations  whose  graphs  intersect  in  a  point 
are  called  simultaneous  linear  equations. 

The  coordinates  of  the  point  of  intersection  of  the  graphs 
of  simultaneous  equations  will  satisfy  each  equation. 

There  are  three  methods  for  finding  algebraically  the 
values  which  will  satisfy  each  of  two  simultaneous  equa- 
tions : 

Elimination  by  addition  or  subtraction; 

Elimination  by  substitution; 

Elimination  by  comparison. 


256 


ALGEBRA  — FIRST  COURSE 


[XVI  §  108 


Problems  in  Loci  and  Simultaneous  Equations. 


Example  1.  If  a  body  moves  along  a  straight  line  3  ft.  per  second, 
how  far  will  it  go  in  10  seconds? 

Solution:  Draw  the  standard  lines  perpendicular  to  one  another. 
Mark  one  t,  on  which  to  count  the  time;  and  the  other  d,  on  which  to 
count  the  distance. 

Construct  a  line  which  is  the  locus  of  distances  passed  over  in  all 
seconds.     By  counting  we  can  quickly  tell  the  distance  passed  over  in 
any  special  number  of  seconds  called  for,  as, 
in  this  case,  10. 

Since  nothing  is  said  about  it,  we  shall 
assume  that  the  body  starts  from  rest. 

The  ratio  of  the  distance  passed  over  dur- 
ing any  number  of  seconds  to  the  number 
of  seconds  is  3.     Algebraically  expressed 

d  =  3t. 

Construct  a  line  through  the  origin  such 
that  the  ratio  of  d  to  i  is  3.  You  can  do 
this  by  counting  to  the  right  1  and  up  3, 
joining  this  point  to  the  origin.  Count  10 
to  the  right  then  upward  until  you  come  to 
this  line,  and  you  have  the  distance  passed 
over  in  1 0  seconds .  So  for  any  other  number 
of  seconds  that  you  choose. 

If  we  regard  the  body  as  not  having 
started  from  rest,  but  as  having  been  in  motion  before  we  became  in- 
terested in  finding  out  about  its  movements,  as  is  frequently  the  case, 
the  question  then  arises,  how  far  away  was  the  body  8  seconds  before, 
or  20  seconds  before,  as  the  case  may  be.  To  find  this  out  count  to 
the  left  8  and  down  until  you  come  to  the  locus  and  you  have  the  answer 
to  the  question. 

Check  answer  by  substituting  8  for  t  in  the  algebraic  equation,  and 
solve  for  d.     Do  you  get  the  same  result? 

The  body  may  be  a  distance  away  from  us  when  we  begin  to  record 
its  movements,  and  we  may  wish  to  know  its  distance  from  us  at  the 
end  of  a  certain  time. 

Example  2.  Suppose  the  body  that  we  were  speaking  of  in  Ex- 
ample 1  is  4  feet  from  us  when  we  begin  to  observe  it. 

For  any  particular  number  of  seconds,  the  distance  as  recorded  by 


XVI,  §  108] 


SUMMARY  —  PROBLEMS 


257 


the  perpendicular  line  will  be  4  feet  less  than  what  it  was  when  we 
considered  it  as  starting  near  us. 

Therefore  the  distance  passed  over  during  any  particular  number  of 
seconds  is  registered  by  a  new  line  parallel  to  the  former  line,  every 
point  in  the  new  line  being  4  units 
below  the  corresponding  point  in  the 
former  line. 

The  quickest  way  to  construct  the 
graph  of  problems  of  this  type  is  to 
first  construct  a  parallel  graph  through 
the  origin  and  from  this  to  construct 
the  required  line. 

Thus,  in  Example  2,  a  body  is  4 
feet  south  of  us  and  is  moving  at  the 
rate  of  3  feet  per  second  northward. 
How  far  from  us  will  it  be  at  the  end 
of  10  seconds? 

The  construction  would  first  be 
made  as  for  the  first  example,  and 
then  as  described  above  for  Exam- 
ple 2. 

1.  How    far   was   the   body 
described  in  Example  2  from  us  5  seconds  before  we  began 
to  count. 

2.  The  algebraic  expression  for  the  thought  of  Example  2  is 

d  =  St-4:. 


d^ 

n^A- 

/ 

Jljr 

-~'i 

-M — 

o/ 

r       ^ 

~-4i 

: 

^T^ 

— 1 

^H-- 

Substitute  the  number  of  seconds  mentioned  in  Exercises 
2  and  3,  solve,  and  compare  with  answers  obtained  from 
graph. 

3.  A  body  is  7J  feet  south  of  us  and  is  traveling  at  the 
rate  of  I  of  a  foot  per  second.  With  reference  to  the  same 
standard  lines  construct  the  graph  of  the  distance  gone  in  t 
seconds. 

How  long  will  it  be  from  the  time  we  began  to  coimt 
until  the  two  bodies  are  together?  How  far  from  us  will 
they  be? 


258  ALGEBRA  —  FIRST  COURSE  Ixvi.  §  108 

4.  Solve  the  equations 

d  =  3  ^  -  4, 
d  =  \t-l\. 

Compare  results  with  those  obtained  from  graph. 

5.  Construct  the  graph  of 

c  =  27rr.  ' 

For  this  work  you  may  use  tt  =  3.1. 

From  the  graph  tell  the  approximate  length  of  the  circum- 
ference whose  radius  is  2.5;  whose  radius  is  3.2. 

6.  Construct  a  graph  which  may  be  used  to  convert  feet 
into  yards.     Explain  its  use. 

7.  Construct  a  graph  by  which  you  may  determine  the 
length  of  the  altitude  of  an  equilateral  triangle  when  you 
have  given  the  side.  Use  Pythagorean  theorem  to  con- 
struct the  length  of  an  irrational  quantity. 

8.  Construct  a  graph  by  which  you  may  determine  the 
momentum  of  different  weights  moving  at  the  rate  of  2 
feet  per  second. 

9.  According  to  physics  the  velocity  which  a  body  falling 
from  rest  acquires  in  t  seconds  is  expressed  by  the  equation 

V  —  gt.     {g  =  32  approximately.) 

Construct  a  graph  by  which  you  can  determine  the  velocity 
at  any  given  second.     Use  a  quite  small  vertical  scale. 

From  the  graph  determine  how  fast  the  body  is  falling  at 
the  beginning  of  the  5th  second;  the  8th  second. 

10.  Suppose  the  body  starts  with  a  velocity  of  3  feet  per 
second,  construct  the  graph  for  determining  the  velocity  at 
the  beginning  of  any  given  second. 

11.  The  distance  passed  over  in  any  one  second  is  ex- 
pressed by  the  equation 

Construct  graph  and  from  it  state  the  distance  passed 
over  during  the  13th  second;  the  17th  second. 


XVI,  §  108]  SUMMARY  —  PROBLEMS  259 

12.  Construct  a  graph  from  which,  by  weighing  a  mass  of 
irregular  shape,  you  can  determine  its  volume  if  you  know 
the  density  of  the  substance  of  which  it  is  composed. 

13.  Two  ladies  arrange  to  attend  a  meeting  together. 
One  lives  8  blocks  north  of  the  hall  where  the  meeting  is 
held,  and  the  other  lives  11  blocks  south.  One  called  up 
the  other  and  said,  ''I  am  just  starting.  You  start  imme- 
diately, and  we  will  meet  and  go  to  the  hall  together." 
One  walked  at  the  rate  of  two  blocks  in  three  minutes,  and 
the  other  at  the  rate  of  three  blocks  in  five  minutes.  De- 
termine by  graph]  the  time  until  they  met  and  the  distance 
and  direction  they  were  from  the  hall. 

SoliUion. 

Let  t  =  number  of  minutes  until  they  meet 

and  d  =  number  of  blocks  the  meeting  place  was 

from  the  hall,  positive  if  north; 
then  S  -{-  d  =  number  of  blocks  1st  walks 

and  11  —  d  =  number  of  blocks  2nd  walks; 

but  1 1  =  number  of  blocks  walked  by  1st 

and  f  t  =  number  of  blocks  walked  by  2nd. 

S  +  d  =  lL     Why? 
And  ll-d=^lL    Why? 

+  d  -  f  <  =  -  8. 
_  d  -  I  <  =  -  n. 
Adding  and  solving:    t  =  15,  no.  of  minutes  until  they  meet. 
Substituting:       8  +  d  =  f  •  15. 

d  =  2,  no.  of  blocks  the  meeting  place  is  north 
of  hall. 
Check  as  in  preceding  problems. 
Compare  answers  with  solution  by  graph. 

14.  Two  cisterns  contain  water,  the  first  to  the  depth  of  4 
feet  and  the  second  to  the  depth  of  7  inches.  Water  is  al- 
lowed to  run  into  them  at  a  uniform  rate  in  each  case,  and 
at  the  end  of  5  minutes  the  water  in  the  first  is  7  inches, 
and  the  other  9  inches.  How  long  will  it  be  before  the  depth 
of  the  water  is  the  same  in  each  cistern,  and  what  is  the 
depth? 


260  ALGEBRA  —  FIRST  COURSE  [xvi,  §  108 

16.  How  long  will  it  be  before  $25  placed  at  5%  simple 
interest  will  amount  to  the  same  as  $28  at  3%  simple  in- 
terest?   What  is  the  amount? 

16.  Solve  Exercises  13,  14,  15  by  algebra,  and  compare 
answers  with  those  obtained  by  graph. 

17.  A  boy  starts  out  on  his  motorcycle  at  the  rate  of  10 
miles  an  hour.  An  hour  later  another  boy  wishing  to  over- 
take him  rides  at  the  rate  of  18  miles  an  hour.  In  what 
time  and  at  what  distance  will  he  come  up  with  him  ?  Solve 
by  both  graph  and  algebra.     Compare  results. 

Solve  the  following  problems  algebraically.* 

18.  Two  boys  were  in  the  habit  of  spending  6  days  each 
spring  in  clearing  out  their  strawberry  patch.  This  year, 
however,  when  it  came  time  to  do  the  work,  one  of  them  was 
ill,  so  that  the  other  one  worked  for  4  days  alone.  Then  the 
first  one  joined,  and  it  too*k  3  days  longer  to  complete  the 
work.  How  long  would  it  have  taken  each  alone  to  do  the 
work? 

19.  A  man  planned  to  plant  an  apple  orchard.  He  en- 
gaged two  men  who  assured  him  that  they  could  finish  the 
work  in  21  days.  It  turned  out,  however,  that  one  did  not 
come  to  work  until  3  days  after  the  other  had  started,  so  that 
it  took  7  days  longer  than  was  planned.  How  many  days 
would  it  have  taken  each  to  do  the  work  alone  ? 

20.  A  reservoir  is  filled  by  means  of  two  pipes.  If  one  is 
open  for  6  hours  and  the  other  is  open  for  7  hours,  the 
reservoir  will  be  filled.  If  one  is  open  for  2  hours  and  the 
other  for  12  hours,  the  reservoir  will  be  filled.  How  long 
would  it  take  each  pipe  alone  to  fill  the  reservoir? 

21.  The  bases  of  two  rectangles  are  7  and  11  feet  re- 
spectively. The  combined  area  of  the  two  is  109  square 
feet.  The  difference  between  the  two  areas  is  67  square 
feet.    What  is  the  area  of  each  ? 

*  If  a  further  extension  of  work  on  loci  is  desired,  the  latter  portion 
of  Part  III  of  Chapter  VI  of  Geometry  may  now  be  studied. 


XVI.  §  108]  SUMMARY  —  PROBLEMS  261 

22.  An  aeroplane  went  66f  miles  in  40  minutes  with  the 
wind  and  53J  miles  in  the  same  time  against  it.  What  was 
the  velocity  of  the  wind?  What  was  the  velocity  of  the 
machine  ? 

23.  The  sum  of  the  two  digits  of  a  number  is  1  more  than 
3  times  the  units  digit.  If  |  be  added  to  ^  of  the  number, 
the  digits  will  be  reversed.     What  is  the  number? 

24.  A  boatman  can  row  12  miles  upstream  in  3  hours, 
and  10  miles  downstream  in  1  hour.  What  is  the  rate  of  the 
stream  and  what  is  the  rate  of  rowing  in  still  water? 

26.  A  number  is  composed  of  two  digits  whose  sum  is  13. 
If  27  be  added  to  the  number,  the  sum  will  be  a  new  number 
with  the  digits  reversed.     What  is  the  number? 

26.  An  ornament  made  of  5  grams  of  brass  and  2  grams  of 
gold  weighs  80.6  grams.  If  it  had  been  made  of  2  grams  of 
brass  and  5  grams  of  gold,  it  would  have  weighed  113.3 
grams.     What  is  the  density  of  brass  and  of  gold? 

27.  A  tank  is  filled  by  two  pipes,  when  both  are  running 
together,  in  9  hours  and  36  minutes.  If  the  smaller  is 
allowed  to  run  for  3  hours,  and  then  both  are  allowed  to  run, 
it  will  take  8  hours  and  24  minutes  after  the  larger  one  is 
opened  to  fill  the  tank.  How  long  would  it  take  each 
running  alone  to  fill  the  tank? 

28.  A  tank  is  filled  by  two  pipes  and  emptied  by  one. 
The  capacity  of  the  smaller  filling  pipe  and  the  emptying 
pipe  is  the  same.  If  the  two  filling  pipes  are  open  and  the 
emptying  pipe  is  closed,  the  tank  is  filled  in  7  hours  and  8t 
minutes.  If  the  smaller  filling  pipe  is  closed  and  the  larger 
filling  pipe  and  the  emptying  pipe  are  open,  the  tank  is  filled 
in  16  hours  and  40  minutes.  How  long  will  it  take  each 
filling  pipe  alone  to  fill  the  tank  ? 

29.  The  boys  of  a  club  were  making  an  assessment  among 
themselves  for  the  expenses  of  a  picnic.  After  fixing  the 
charge  for  each  boy,  4  said  that  they  could  not  go,  so  it  was 
necessary  to  tax  the  rest  10  cents  more  per  boy.    Later  4 


262  ALGEBRA  —  FIRST  COURSE  Ixvi.  §108 

more  decided  not  to  go.  It  was  necessary  to  make  another 
increase  in  taxes  per  boy  of  15  cents.  How  many  boys 
were  there  in  the  club,  and  how  much  was  the  original 
assessment  per  member? 

30.  If  a  rectangle  of  given  area  had  10  feet  added  to  its 
base  and  5  feet  subtracted  from  its  altitude  its  area  would 
not  be  changed.  Again  if  it  had  5  feet  subtracted  from  its 
base  and  4  feet  added  to  its  altitude,  its  area  would  not  be 
changed.  What  is  the  length  of  the  base  and  the  altitude 
of  the  rectangles? 

31.  If  19 J  pounds  of  gold  and  10|  pounds  of  silver  each 
lose  one  pound  when  weighed  in  water,  how  much  silver  and 
how  much  gold  is  there  in  a  mass  which  weighs  20  pounds 
and  loses  IJ  pounds  when  weighed  in  water? 

32.  There  are  two  angles  the  sum  of  whose  complements 
is  75 1  degrees  and  the  difference  of  whose  supplements  Is 
28 J  degrees.     What  is  the  number  of  degrees  in  each  angle? 

33.  There  are  two  angles  such  that  the  supplement  of 
the  first  is  8  degrees  less  than  three  times  the  complement 
of  the  second,  and  the  complement  of  the  first  is  20^  degrees 
more  than  J  of  the  supplement  of  the  second.  What  is  the 
size  of  each  angle? 


CHAPTER  XVII 
EXPONENTS   AND   RADICALS  * 

109.  Definitions.  We  have  already  used  the  following 
abbreviations. 

a^  stands  for  a  »  a  '  a;  a^  stands  for  a*  a*  a*  a,  etc. 
Likewise : 

a*"  stands  for  a  •  a  •  a  •  .  .  .  to  m  factors,  m  being  a  positive 
integer; 

a"  stands  f or  a  •  a  •  a  •  .  .  .  to  n  factors,  n  being  a  positive 
integer. 

110.  To  multiply  a^  by  a'\  By  direct  multiplication  we 
see  that: 


'  a  '  a'  a  =  a'  =  a^+^; 

and  so  in  general, 

a^  Xa""  =  a  '  a*  a*  .  . 

.  to  m  factors  X 

a'  a*  a*  .  . 

.  to  n  factors 

=  a  '  a*  a' a* 

.  .  .  to  m  +  n  factors 

=  a"'+''. 

So  we  have  our  first  rule. 

Rule  I.     To  multiply  a*"  by  a**,  add  the  exponents;  the  prod- 
uct is  0"^+". 

As  an  equation: 

*  For  closer  correlation  it  would  be  well  to  take  up  Chapter  VII  of 
Geometry  before  going  on  with  the  study  of  exponents. 

263 


264  ALGEBRA  —  FIRST  COURSE  Ixvii,  §  in 

Exercises.     Apply  this  rule  to  the  following  products: 

1.  22.23.  5.  r3.r'«.  9.  (-3c)3(-3c)^ 

2.  34.32.  6.  t^'t\  10.  aP.a«. 

3.  4  .  43.  7.  (-  6)2  (-  6)3.  11.  ap  -a^'  a\ 

4.  a^ .  a*.  8.  (2  aY  (2  a)\  12.  x^  -  x^  -  xK 

111.   To  multiply  d^  by  6***.     By  pairing  off  the  a's  and 

6's  we  can  easily  see  that  the  following  results  are  correct. 

a2  X  62  =  a  .  a  X  6  .  6  =  a6  .  a6  =  (a6)2.     Why? 

a3  X  63  =  a  .  a  .  a  X  6  .  6  .  6  =  a6  .  a6  .  a6=  (a6)3.     Why? 

So  in  general, 

a""  xh"'  =  ah  '  ab  *  ah  '  .  .  .  to  m  factors  =  (a6)'". 

Rule  II.     The  product  of  a""  hy  6"*  is  equal  to  {ah)"'. 

That  is:  a*"  •  6*"  =  (a6)'". 

In  the  same  way  we  can  show  that 


Do  this,  and  state  the  rule  in  words. 

Example  1. 

23 .  33  =  (2  .  3)3  =  63  =  216. 

Example  2. 


63      /6\3 
33      1,3/    " 

=  23 

=  8. 

Exercises.     Reduce  the  following 

expressions: 

1.  33.43. 

2.  2^.5*. 

'    4^ 

-I- 

3.   63.58. 

12^ 

„  1^' 

4.   4«.5«. 

«•   IT- 

11.  303' 

6.    2^.3^.5^ 

Q    103 

,0    27* 

6.   73.53.2'. 

^-    23- 

12.   81. 

XVII.  §1121  EXPONENTS  AND  RADICALS  265 

112.   To  divide  a"*  by  a'\    By  cancellation  we  get  such 
equations  as  the  following: 

a^      a*  a*  a*  a  -  a        ,         ^„ 


=  a" 

'•  a'  a 

a' 


a^  a'  a 

,7 


So  in  general  we  have  the  rule, 

^  =  a'"-",   provided  m  is  greater  than  n. 

Here  we  subtract  the  exponent  of  the  denominator  from 
the  exponent  of  the  numerator. 

Now  what  is  the  result  when  m  is  less  than  n,  that  is, 
when  there  are  fewer  a's  in  the  numerator  of  the  fraction 
than  in  the  denominator?     Let  us  look  at  some  examples. 


aP-              a- a             .1 

1 

a^      a  'a-  a'  a*  a      a'  a  '  a 

a' 

Likewise: 

a3       1           a'       I 

Now  we  shall  use  the  symbol  a~"  to  stand  for  — ;  for 

a"' 


example : 

1  4         1             ^ 

— ;  a~^  =  — : :  and  so  on. 

/IT'S    '  /T*   ' 


a 
Then  we  have 


-3_ 


or      a** 
a^       1 


a 


=  a-' 


^  =  1  =    -3 


We  see  again  that  we  get  the  result  by  subtracting  the 
exponent  of  the  denominator  from  the  exponent  of  the 
numerator. 


266  ALGEBRA  —  FIRST  COURSE  [xvii.  §  ii4 

Rule  III.  To  divide  o/^  by  a"",  subtract  the  exponent  of  the 
divisor  from  the  exponent  of  the  dividend;  the  quotient  is  a"*"", 
whether  m  is  greater  or  less  than  n. 

Exercises.     Apply  this  rule  to  the  following  quotients: 

1.4 


a'^ 

2    — ■ 

a' 


9. 


a^'a° 
a^  •  a' 


3.  S-                - 

24 

2^* 

3^ 

3^* 

7     (2  a? 
{2  ay 
o     (3  6)^ 

{sby 

(First  reduce  numerator  and  denominator  by 
Rule  I.) 

^  ^     b'^  '¥                  ^  ^     x^ '  x^ '  x^ 

10.    ^^-  11.    f,r^-  12. 


a^ '  a^  '   b^  'b"^  '    x^ '  x^  *  x^ 

113.  The  Meaning  of  a^.  Suppose  that  in  Rule  III  n  is 
equal  to  m;  for  example,  suppose  m  =  3  and  n  =  3.  Fol- 
lowing the  rule,  we  have 

a^ 

^3 

But  -;  =  1;  therefore  a^  =  1. 

a^ 

In  the  same  way 
a^ 
a"^ 

But  —  =  1;  therefore  a^  —  1. 

a"" 

That  is,  we  must  regard  a°  as  equal  to  1,  if  Rule  III  is  to 
apply  to  cases  where  m  and  n  are  equal.     So  we  have 

««  =  1. 

114.  The  Meaning  of  (a**')*'.  The  expression  (a*")"  stands 
for  the  quantity 

^m  .  ^m  .  ^TO  .  ^  ^  ^  iq  ^  factors. 

But  a^  =  a  *  a*  a  '  .  .  .  to  m  factors. 


XVII.  §115]  EXPONENTS  AND  RADICALS  267 

Therefore  (a"*)"  =  a*  a  *  a*  a'  .  .  .  to  m  X  n  factors; 
that  is,  (a"^)"  =  a"»  "  =  a"*". 

Rule  IV.     To  raise  the  expression  a""  to  the  nth  power,  mul- 
tiply its  exponent  by  n. 
As  an  equation, 

The  same  rule  is  true  when  n  is  negative,  or  when  m  is 
negative,  or  when  both  m  and  n  are  negative.  This  will  be 
seen  from  the  following  examples. 

1.  To  show  that  (a4)-3  =  a^^^c-^)  =  a-'\ 

(n^)-^  =  = =  J-  =  0-12 

^""^         a' -a' -a'      a'^      ""     ' 

2.  To  show  that  {a-^y  =  a-^""^  =  a-'^. 

^      ^  a"^   a^   a^        a^^ 

3.  To  show  that  (a-^)-^  =  a-^^^-^)  =  a^\ 

115,  Summary  of  Results.  Collecting  our  results  we 
have: 

Rule  I.  a*^  .  a»»  =  «»**+»*. 

Rule  II.       a-.  6- =(«&)-;       ^  =  Q    • 

Rule  m.    a"*  -^  a"  =  «♦"-»». 
Rule  IV.         (O**  =  a»"»». 

Also:  «-"  =  — ^  ;  a®  =  1. 

Here  m  and  ti  may  be  either  positive  or  negative. 

Examples. 

1.   Multiply  2  a362c4  by  3  a^h^cK 
Multiplying  together  the  like  factors  we  have 
2  a^hH"^  X  3  a%h^  =  6  a^lP&. 


268  ALGEBRA  —  FIRST  COURSE  [xvii.  §  115 

Similarly, 


3.  (a;-2  +  2/"')  (^"'  -  2/"')  =  (^-')'  -  {y-''Y  =  a:-'  -  y-'- 


-         /,-2  _  ^-2           -    V- 

=  /i-^  +  h-%-^  +  /b-S 

111 

ki  +  /i2/b2  +  /i4 

5.   Simplify  [(-  2a;3)-4]6  -f-  [64  (x^y]-^ 
Starting  with  the  dividend: 

{-2x')-'  =  (-2)-4.a;-i2. 

Then  [{-2x')-'f  =  [(-2)-''x-^^f=  i-2)-^''x--'^  =  2-^'x-''\ 

Simplifying  the  divisor: 

[64  (x^y]-'  =  [2^  ^]-'  =  2-3«  x-^. 
Therefore 

64 

[(-2  x^y^^f  ^  [64  (x2)4]-5  =  2-24a;-72  ^  2-3'^a;-^o  =  2«a;-32  =  -^. 

Exercises.     Simplify  the   following;  write  answers  with 
negative  exponents,  and  also  without  negative  exponents. 

1.  xy.o^V-  13.    (22)-2.  22.  (a" +  6^)2. 

2.  a-36-2 .  a-26-3.  14.  [(22)-2]2.  23.  (x"*  -  y'^y, 

3.  a-4  -J-  a-\  16.   4"*  •  2^.  24.  (p""  +  g~")^ 

4.  5m^  -T-  52  m-2.                 6^  25.  x-^  (1  +  ^2). 
6.   p0g-2.2-2§^                 24.3^*  26.  x3(l-a:-3). 

6.  c^d^  -4-  3-2^3.  8^  »  4-3  27.    (r2s-2  -  1)2. 

7.  2rs2.s-2.  27.2-^'  a2"^-2 

8.  3a-l-^3-2a-^  16a-26-3  ^°'  a2"»-3 

9.  (2 w) 2.  (3^3). 2^-5/®-  2^a-3  6-4*  a2^b»+^  > 

10.  xo+i/'.  19.    (a2  +  62)2.        ^^'   a"»62"-i' 

11.  Q^  -xf^.  20.    (x3  -  2/3)2.  ^i-^z;i+" 

12.  53(5x-i)-3.  21.    (w-^  +  z;-^'.  w"+i2;"-i' 


36.    (l-a:2). 

38. 

36.    (1  -  x-2). 

39. 

37.    (a2-62). 

40. 

41. 

(X3  -  2,3). 

42. 

(X-'  -  r')- 

43. 

(X-'  +  y-'). 

XVII,  §1161  EXPONENTS  AND  RADICALS  269 

31.  The  edge  of  one  cube  is  x  and  of  another  cube  2  x\ 
what  is  the  ratio  of  their  volumes  ? 

32.  What  is  the  ratio  of  the  volume  of  two  cubes  whose 
edges  are  2  x  and  3  x  respectively  ? 

33.  What  is  the  ratio  of  the  volume  of  a  sphere  whose 
radius  is  r  to  a  cube  whose  edge  is  r? 

.34.  What  is  the  ratio  of  the  volume  of  a  sphere  of  radius  r 
to  the  volume  of  a  cylinder  of  radius  r  and  height  h  ? 
Factor  the  following: 

(a-2  -  6-2). 

116.  Fractional  Exponents. 

Example  1.  What  is  the  side  of  a  square  whose  area  is  2? 
We  know  that  the  answer  must  be  a  number  which  when 
multiplied  by  itself  will  give  2.  We  express  this  number  by 
the  symbol  2^  (or  by  V2). 

Similarly  if  the  area  of  a  square  is  a,  its  side  is  expressed 
by  a^.  Here  a^  is  a  symbol  used  to  indicate  the  number 
whose  square  is  a;  as  an  equation, 

Rule  I  applies  here;  for  using  this  rule, 
a^  '  a^  =  a2   2  z=  a^  =  a. 

Example  2.  What  is  the  edge  of  a  cube  whose  volume  is 
2?  This  must  be  a  number  which  when  multiplied  by 
itself  three  times  will  give  2.  We  express  this  number  by 
the  symbol  2^  (or  by  \/2).     It  is  called  the  cube  root  of  2. 

If  the  volume  of  the  cube  is  a,  its  edge  is  expressed  by  the 

symbol  a^;  it  is  called  the  cube  root  of  a.    We  must  then 

have 

111 
as .  a3 .  a3  =  a. 

Rule  I  again  applies;  for  it  gives 

as .  a*  •  o^  =  a^^  3^^  =  a*  =  a. 


270  ALGEBRA  —  FIRST  COURSE  [xvil.§li6 

Following  out  this  idea,  we  define  the  symbol  a"  as  follows, 
n  being  a  positive  integer,     i 

Definition.  The  symbol  a""  (or  \/a)  expresses  that  number 
which  when  multiplied  by  itself  n  times  will  give  a;  it  is 
called  the  nth  root  of  a.     That  is, 

111 
a"" '  a"" '  a^  *  .  .  .  to  n  factors  =  a. 

This  is  in  accordance  with  Rule  I;  for 


ol"  -a"^  *  aJ" '  .  ,  .  to  n  factors 
a' 


— I (--4-  .  .  .  to  n  terms 

nV.    n    n 


1 

=  a    ^  =  a'  =  a. 
We  next  obtain  a  meaning  for  a  symbol  such  as  aK     In 
order  to  make  Rule  I  apply  to  such  cases  we  think  of  a^  as 
follows : 

111  I-4.I4.I  8 

That  is,  we  regard  a^  as  the  cube  of  a^. 

m 

In  exactly  the  same  way  we  regard  a"  as  the  mth  power 
1 

of  a". 

We  may  also,  using  Rule  IV,  regard  a^  as  follows: 

That  is,  a^  is  the  same  as  the  square  root  of  a^. 
Similarly,  if  Rule  IV  is  used, 

1  1  m 

mx-  — 

{a^Y  =  a     "  =  a". 

tn 

So  we  regard  d^  as  the  nth  root  of  a"». 

m 

We  may  thus  think  of  a"  as  a  symbol  representing  the  mth 
power  of  the  nth  root  of  a,  or  as  representing  the  nth  root  of 
the  mth  power  of  a. 

In  algebraic  language  this  is: 

m        /     \\m  1 


XVII.  §116]  EXPONENTS  AND  RADICALS  271 

Exercises.   Express  each  of  the  following  in  the  two  forms 
just  given;  state  each  result  in  words. 

1.   3l  5.   27i  9.   ai  13.   xl 


2.  23.  6.   81*.  10.   a^.  14.   y^. 

3.  5i  7.    (i)i  11.   6i  15.   cl 

'    4.   32i  8..(TJ^)i  12.   6i  16.    (a +  6)1 

17.  If  the  volume  of  a  cube  is  8,  show  that  the  area  of  one 
of  its  faces  is  8^  =  4.  If  the  volume  is  Vj  show  that  the 
area  of  one  face  is  v^. 

18.  If  the  area  of  one  face  of  a  cube  is  9,  show  that  its 
volume  is  9^  =  27.  If  the  area  of  one  face  is  a,  show  that 
the  volume  is  a^. 

A  negative  fractional  exponent  we  define  in  the  same  man- 
ner as  a  negative  integral  exponent;  that  is, 

1 


Thus: 


It  is  easy  to  see  that  Rule  II  applies  also  to  the  case  of 

fractional  exponents.     By  Rule  II  we  would  have 

1        1 
1        1  1 


the  symbol  a 

"  means 

X 

m 

a" 

a 

-\ 

1 

1 

a-t  = 

1 

1 

fi' 

-I 

1 
16^ 

1 

1 
I      2z 

1 

8* 

.u 

(16^)^ 

{obY  =  a^'b'',     and     f  H   =  -j 


But  both  of  these  equations  are  correct;  for  raising  both 
sides  of  each  equation  to  the  Tith  power,  these  equations 
become 

ah  =  ab,  and  r  =  r* 

0         0 

Example,     ^bi  =  (54)^  =  (27  •  2)  ^  =  27*  •  2^  =  3  ^2. 


272  ALGEBRA  —  FIRST  COURSE  [xvii,§ll8 

Exercises.     Reduce  each  of  the  following  as  in  the  last 
example. 

1.  \^TQ.  3.    32*.  6.    </40.  7.    24^ 

2.  -V^Sl.  4.   64^.  6.    v'SO.  8.    108^ 

Let  us  now  examine  a  product  like  a"'*a''.    We  can  easily 
show  that  (work  this  through,  taking  m  =  2  and  n  =  3.) 


For 


also, 
Then 


1     1 

= 

^4 

1 

= 

a^  =  \ 

fjmn 

n 

} 

1 
a" 

= 

m 
gmn  = 

(  -^ 

r-, 

gm     n 

= 

m-\-n 
amn     = 

/     1  \m+ii 

-   -    (  -Y  (  -V 

(< 


\m+n 


^  amn 

-!■+-' 

—  gm     n 


Show  in  a  similar  manner  that  Rule  III  applies  to  frac- 
tional exponents;  that  is,  that 

i        1        1_1 

(jm  ^  gn  —  gtn     n^ 

Again  work  out  the  case  where  m  =  2  and  n  =  3. 

Finally  we  can  show  that  Rules  I  and  III  apply  in  general 
as  follows : 

Example.    Show  that 

at .  a^  =  a^"^7. 

Proof.    The  common  denominator  of  the  exponents  is  15. 
Raise  both  sides  of  the  equation  to  the  15th  power.    We  get 


XVII,  §116]  EXPONENTS  AND  RADICALS  273 

Simplifying  we  get 


ai« 

.ai2  : 

=  a  10+12^ 

bich  is  a  true  equation. 
Prove  in  the  same  way 

that 

atn 

rJ: 

=  J-K 

Similarly  we  show  that 

p 

r 

a'  = 

E+r. 

E 

r 

E-t 

Prove  these  by  raising  both  sides  of  each  equation  to  the 
q  •  sth  power.  Does  this  proof  require  that  the  letters 
Vj  Q)  f}  ^)  stand  for  positive  numbers?  Try  a  numerical 
example. 

Finally  let  us  consider  Rule  IV,  when  m  and  n  are  frac- 
tional numbers. 

Example.    Verify  that  (a^)^  =  a^^^=  a\ 

Raising  to  the  6th  power, 

{a^y  =  a^     (Why?) 
Taking  the  fourth  root, 

af  =  ai     (Why?) 

Similarly  we  can  verify  Rule  IV  when  m  and  n  are  any 
two  fractional  numbers. 

So  we  see  that  the  rules  on  p.  267  apply  to  fractional  as 
well  as  to  integral  exponents. 

Exercises.  Simplify  the  following  expressions,  giving  the 
reason  for  each  step. 

1.  4^  •  4i  4.   a^  •  a*.  7.   s~^  -^  s^^ . 

2.  8^.8-i  5.   x^'X^.  8.    (6^)i 

3.  16^-^8i  6.  c^'C^-c^K  9.    (m-t)i. 

10.  Show  that  a?  -  6t  =  (ai  +  h^)  (a^  -  fei). 

11.  Factor  x^  —  y^.  12.   Factor  u^  —  v^. 


274  ALGEBRA  —  FIRST  COURSE  ixvii.  §  118 

13.  Multiply  a^  +  b^  by  a^  -  a^b^  +  bK 

14.  Multiply  x^  —  i/3  by  x^  +  x^y^  +  yK 

15.  Divide  x  —  y  hy  x^  —  y^. 

16.  Divide  u^  —  v"^  by  Vu  —  Vv. 

17.  Solve  for  a;:  a;^  —  3  a;^  +  2  =  0.  (Regard  x^  as  un- 
known.) 

18.  Solve  ioT  x:  x^  -\-  7  x^  -  S  =  0. 

19.  Is  \/a2  +  62  equal  to  a  +  6?  Is  \^a^  +  b^  equal  to 
a  +  6? 

117.  Irrational  Numbers  or  Surds. 

Definitions.  Numbers  such  as  3  V2,  4  v^,  etc.,  which 
involve  an  indicated  root  that  can  not  be  exactly  found,  are 
called  irrational  numbers  or  surds.     See  p.  176. 

The  number  under  the  radical  sign  is  called  the  base  of 
the  surd,  and  the  number  indicating  the  root  to  be  extracted 
is  called  the  index  of  the  surd. 

Surds  expressed  by  the  same  base  and  index  are  called 
similar  surds. 

Thus:  4  \^,  7  v^5,   —  3  ^  are  similar  surds. 

Also:  "V^  and  v^  are  dissimilar  surds. 

118.  Examples  Involving  Surds. 

Example  1.     Reduce  V48,  V75,  and  V27  to  similar  surds. 
These  can  all  be  expressed  as  multiples  of  Vs. 
For:  n/48  =  VlQ^s  =  4  Vs.     (Why?) 

V75  =  V25T3  =  5  V3;   V27  =  V9T3  =  3  V3. 
Example  2.     Simplify  V48  +  V75  -  \/27. 

Vis  +  V75  -  V27  =  4  V3  +  5  V3  -  3  V3  =  6  Vs. 
Example  3.     Simplify  16^  -  54^  +  128l 

16l  =  8i  •  2^  =  4  .  2^, 

54!  =  27^  •  2^  =  9  .  2t, 
128^  =  64^  .  2^  =  16  .  2I 

let  _  54?  +  128^  =  4  .  2^  -  9  .  2^  +  16  •  25  =  11 .  2i 


XVII,  §118]  IRRATIONAL  NUMBERS  275 

Rule.  To  combine  surds  by  addition  or  subtraction,  they 
must  first  be  reduced  to  their  simplest  forms.  Then  if  they  are 
similar  they  may  be  combined;  not  otherwise. 

Exercises.     Simplify  the  following. 
1.   5  VS  +  3  Vl8.  6.   2  .  63^  +  5  .  28^  -  175^ 

-     2.   3^^-2v^.  6.   3.44^  +  i.99^+i.275i 

3.  i  a/125  +  J  V20.  7.    (i)^  -  in)^  -  {i^)K 

4.  V|  +  V3.  8.   a(18a36)^-(a2-62)(2a6)i 

9.   3v^-5v^l6-2v/432. 

10.    (t\)^-(/t)^.  

11.   2ab<^-2\^o}W.      12.   5  m  a^27w^  -  3  v'^Sm'^ 

V3 
Example  4.     Find  the  value  of  j=  to  cwo  decimals. 

Rationalizing  the  denominator: 

Vs     _     Vs     ^2  + V3 
2-V3~2-V3*2  +  V3 

Now  replace  Vs  by  its  approximate  value,  1.73; 

V3 


2-  V3 


=  2-1.73  +  3  =6.46. 


Exercises.     Calculate  each  of  the  following  to  two  decimal 
places. 

1.     2  .  ^_     1  .  3_   .  /8 


Vs  V5  ^5  V45 

2  -  2  V3  .  1 


6.    7='  7. 


7-Vq  3  +  \/3  V27-VI8 

2  +  ^5^  10  VTl  +  \/7 

2-V5'  *   V3  +  V8*  *   VII-V7* 


276  ALGEBRA  —  FIRST  COURSE  [xvii.§ii8 

Rationalize  the  denominators  in  the  following: 

11               ^                      13           ^                           15       1  +  ^ 
11.     — ;;= 7=:*  lo. T  lO.     r' 

.„    a^  +  b^  Vm  + V2  Vx^  -  2 

12«       1        —J-  •  14.        , —  ,_  •         16.    7=.  • 

a^  _  5§  Vm  -  V2  2  +  Va;^ 

Example  5.     Reduce  V2,  "A?^,  v^  to  surds  with  the  same 
index. 
Write  each  surd  with  a  fractional  exponent,  thus 

2^     5^     3i 
Reduce  these  exponents  to  the  least  common  denominator : 

2^=2t^;     5^  =  51^;     3^  =  3^'^. 
But  2t'2  =  W«  =  V64; 

5^  =  ^^5^  =  7625; 
3^  =  W^  =  'V^. 
So  the  three  given  surds  are  equal  respectively  to 

■•V64,      n/625,     727. 
These  have  the  common  index  12. 

Exercises.     Reduce  the  following  to  surds  with  the  same 
index. 

1.  V3,  </2.  3.    </%  a73.  5.    \/|,  v^,  v^. 

2.  </l,  VI.  4.    V3,  ^,  .\^.       6.    A^i  Vi,  ^|. 

7.    VT^,  V^a,  v^.  8.    </'M,  <^\  </d>. 

Example  6.     Multiply  Wa}  by  ^v/8a"^ 
First  reduce  to  same  index: 

VSo^  =  V(8a3)^  =  V'S^T^  =  \/29T^. 
Then  \/4a2  •  <J%a^  =  Wo^  •  "^29^ 

=  V2i^^^ 
=  2a\/32c?. 


XVII.  §119]  IRRATIONAL  NUMBERS  277 

Similarly  in  division  of  surds  it  is  usually  best  to  reduce 
to  the  same  index. 

Exercises.    Perform  the  following  operations,  reducing  all 
results  to  simplest  form. 

v^o^.V^^  7.  </^x^'</2x\ 
Vm  '  y/m.  8.  a  \^  •  VSa. 
</7^'</x\      9.    ^^^506  .  VIOo^P. 


1.  ^-^/z. 

4. 

2.   ^.^. 

6. 

3.    Vi-^i. 

6. 

] 

..     <^' 

12.    -^y='  14. 

V  6^  V  u° 


11.    — ^-  13.    -i7:=^-  16.    -17=^- 

Vx'  </'m}  </2a 

119.  Equations  with  Irrational  Terms.  Occasionally 
equations  arise  which  contain  irrational  terms.  Several 
times  our  equations  have  contained  one  irrational  term. 
You  saw  readily  that  these  could  be  easily  handled  by 
writing  the  irrational  term  as  one  member  of  the  equation, 
and  the  rational  terms  as  the  other  member;  you  could  then 
easily  rid  your  equation  of  the  irrational  term  by  squaring 
both  members. 

When  an  equation  contains  two  irrational  terms,  it  is 
customary  to  write  one  of  the  irrational  terms  as  one  mem- 
ber and  the  other  irrational  term  together  with  the  rational 
terms  as  the  other  member.  Squaring  both  members  of 
the  equation  will  rid  the  equation  of  one  of  the  irrational 
terms.  After  this  the  equation  can  be  rid  of  the  other 
irrational  term  as  above  described.  The  following  illustra- 
tions will  show  these  processes. 

Example  1.     Given  {x  +  3)^  =  7. 
Squaring:  a;  +  3  =  49;  * 

X  =  46. 
Check  by  substituting  in  the  first  member  of  the  equation. 
(46  -f-  3)^  =  49^  =  7, 
as  the  problem  states  that  it  should. 


278  ALGEBRA  —  FIRST  COURSE  ixvii.§ii9 

Example  2.     Given 

(x  +  3)^  +  (x  -  6)^  =  9, 

(x  +  3)^  =  9  -  (x  -  6)^ 
Squaring:  x  +  3  =  x  +  75  -  18  (x  -  6)^ 

-  72  =  -  18  (x  -  6)^ 
4  =  (a;  -  6)s 
16  =  X  -  6, 
X  =  22. 

Check  by  substituting  in  the  first  member  of  the  equation. 
(22  +  3)^  +  (22  -  6)^=  5  +  4  =  9, 
as  the  given  equation  calls  for. 

Example  3.     Given 

V2x  +  2  -  V4x  -  3  +  Vx  -  6  =  0. 
V2x  +  2  =  \/4x-  3  -  Vx-  6. 
Squaring:        2x  +  2  =  5x-9-2  V4x-  3  Vx  -  6, 

-3x  +  ll  =  -2V4x-3Vx-6. 
Squaring:  9  x^  -  66  x  +  121  =  16  x^  -  108  x  +  72. 
7  x2  -  42  X  -  49  =  0, 
x2  -    6  X  -    7  =  0. 

Solve  and  finish  as  in  past  examples  of  this  kind.     Check 
as  in  preceding  examples. 

Exercises.     Solve  the  following;   check  all  answers. 
1.    (2x  +  9)^=5.  2.    (5x-9)^  =  6. 

3.  (2x-6)^-  (5x-6)^  =-3. 

4.  V3x-2  -  Vx-'l  -  V^'^T^  =  0. 

^     (2x+l)^_3*  ^      5x-9        .  _V6x-S 

6.    — , 1  —  ^ • 


(3x  +  4)^      4  V5x-\-S 

V3 X  4-1  +  V3x 
V3x  +  1  -  VSx 


=  4. 


XVII.  §  120]  SUMMARY  —  EXERCISES  279 

120.  Summary  and  Exercises  for  Review. 

The  symbol  a"^  when  m  is  a  positive  integer,  means  the 
product  a'  a'  a*  .  .  .  to  m  factors. 

The  symbol  a"*,  when  m  is  a  positive  integer  means  that 

number  whose  mth  power  is  a;  that  is,  a""*  a""*  a"^'  .  .  .   to 
m  factors  is  a. 

1  -^  1 

a""*  means  — - ;         a  "»  means  —r] 

or 
a  "  means  —  ;         m  general,  a-'  =  —  • 


m 


a' 


Rules.  These  rules  apply  whether  the  exponents  are  in- 
tegral or  fractional,  positive  or  negative. 

aP  -T-  a*^  =  aP-«. 

apfcp  =  (ahy. 
State  these  rules  in  words. 

Surds  (irrational  numbers)  are  numbers  involving  indi- 
cated roots  that  can  not  be  exactly  found.  Example,  \^5. 
Here  5  is  called  the  base  and  3  the  index. 

Surds  expressible  by  the  same  base  and  index  are  similar. 

To  add  or  subtract  surds,  they  must  be  similar. 

In  working  with  surds,  it  is  often  best  to  reduce  radical 
forms  to  fractional  exponents.  Then  apply  the  rules  given 
above. 

Miscellaneous  Exercises  for  Review. 

Simplify  the  following,,  and  write  each  result  in  the  form 
of  a  single  radical. 


;o 

ALGEBRA  —  FIRST  COURSE 

[XVII.  §  120 

1. 

(a)  3^  .  2^; 

(b) 

8^  .  163;       (c) 

aibi; 

{d)  16^x1 

2. 

(a)  3(2^c^); 

(6)  4(at6i); 

(c)  2sa:M. 

3. 

(a)  (2a^)s 

(6)  5(7^at)l; 

(c)  (m^n^)l 

4. 

(a)  [(a;t)i]l; 

(&)  [(2  6)¥; 

(c)  (c^dt)l 

/  X  5a^ 

,,,  7  mi 

,  ,  2  a^d^ 

6. 

2a^ 

^'^  3al6f 

In  the  following  perform  indicated  operations  wherever 
possible  and  reduce  all  results  to  their  simplest  forms. 
State  rules  and  principles  used. 

6.  {ai  +  h^y.  12.  (Vr  +  ^s){Vr-<rs), 

7.  {a^  +  b^y.  13.  (Vx  +  Vy  +  Viy. 

8.  {a^  +  h^)  {a^  -h^).  14.  {m  +  n)Hm-  n)f. 

9.  {s  -t)  -^  {s^  -t^).  16.  (m^ +  n^)  (m~2  4-n"^). 


10 


11.    (^p2  -  V^)l  17.    Va  +  6  •  Va  -  b, 

18.  (m  +  Vmn  +  n)  (m  —  Vmn  +  n). 

19.  (2  w^  +  3  Vm)  (3  m^  -  2  Vm). 


20. 
21. 
22. 


gl  -2a^b\ 
ah^  -2b' 
x^  —  y^ 

XZ  —  yi 

(Vm  —  Vn)' 


23. 

a* 

-6^ 

oi- 

•a^fti 

24. 

X  - 

y 

OR 

x  +  sVx 

+  2 

ffini.      X  +  3  Vi  H-  2  =  (VS  +  l)  (Vi  +  2). 

2/3  —  41/3—4  "V^r         Vr 

■   yl  _  5  yi  +  6  '   2  Vr      Z-i/r 


m 


^-8m^  +  15  Va;  +  Vi/     Vx-Vy 


XVII.  §  120]                 SUMMARY  —  EXERCISES  281 

28.    ;=— 34.    —^ -T  —  2. 


r 


2  Vrs  —  15  s  x^  +  x^ 


v^a;2  -  3  v^  +  2 


m'  m- 

30-  z:^ — z:=^-r 


v^^2  _  1  m^  —  m~2      m^  +  mr^ 


pf-plgf-20gt  1+3S-3      1-38-3 

pi  +  pf^f  -  30 5^*  ^^'  1-Ss-'      1+3 s-^' 

,^     Va  + Vb      Va-Vh  1  ,       2t/-i 

31.  7= 7=      *      37.  :     ;  r  "T  ;i  j 

3  Va  4  Va       i^"'  +  2/"'      a;-^  -  ^/"^ 


39. 


2(a-3  +  6-3)   '  a-«-6-« 


x^  {pr^  +  a;'-)      or'  (a"'"  +  a^O 


40.^:-. 2^^+        ^ 


41. 


m^  +  1      m"  +  2      m"  +  3 
3  5  2a:3_7 


7?      2^  -\      4a;6_  1 

_     2    ,         3  2a;^-3 

42.    -T  H 1 ^ 

x^      l-2x^      4a:3-l 

43.  7  \/2  -  VTS.  49.  (V2  +  V7)  (V2  -  V?). 

44.  9>/5-2v^.  50.  (l  +  V^2)l 

45.  V75+V48-v1l7.  51.  (v/9-v^)l 

46.  5v^  +  2v/32-v^T08.  52.  VI+VfvT^Tv?. 

47.  V2+3V32  +  J\/l28.  53.  -x/o  +  VTlVe  -  VH. 

48.  3  A^^S- 5^/48+^^243.  54.  (5+2\/3)  (3- 5\/3). 

55.    (^-2>J/4)(4^  +  ^). 

56.  V6.Vl2.V72.  69.    Vs^ .  Vs^  •  V5«. 

57.  V3  .  Vl8.  60.    Vg  .  Vsl .  V729. 

58.  VsI.V^^^.  61.    V3(V8r- V243). 
Solve  the  following  equations: 

62.  5  x^  -  3  a;3  -  2  =  0.       64.   3  m^  -  7  w^  -  6  =  0. 

63.  9  2/^  +  15  2/*  -  6  =  0.      65.    7  r  -  9  r"  -  10  =  0. 

66.   2t^-2-  13^^-1+  15  =0. 


CHAPTER  XVIII 
BINOMIAL   THEOREM 

121.  The  Binomial  Theorem.  By  multiplying  find  the 
values  of  the  following. 

(a  +  hy;  (a  +  by;  {a  +  bY;  {a  +  by. 
.  Examine  the  exponent  of  the  first  term  in  each  expansion. 
Is  it  the  same  as  the  exponent  of  the  binomial?    What 
would  be  the  first  term  of  (a  +  by  ? 

Examine  the  exponents  of  the  successive  term  of  each  of 
the  expansions.  What  do  you  find  of  interest  about  them  ? 
What  will  be  the  exponents  of  a  and  b  in  each  of  the  suc- 
cessive terms  of  (a  +  by  ? 

Examine  the  coefficient  of  the  first  term  of  each  expansion. 
What  is  it?  What  is  the  coefficient  of  the  first  term  of 
{a  +  by? 

What  do  you  find  of  interest  about  the  coefficient  of  the 

second  term  of  each  expansion?     What  is  the  second  term 

of  (a +  6)7? 

3  •  2 
Will  z — ^  give  you  the  coefficient  of  the  third  term  of  the 

4  •  3 
expansion  of  (a  +  6)^?     Will  :; — ^  give  the  coefficient  of  the 

5  •  4 
third  term  of  the  expansion  of  (a-\-by?     Will  y—x   give 

the  coefficient  of  the  third  term  of  the  expansion  oi  (a  -\-  by? 
Write  by  this  rule  the  coefficient  of  the  third  term  of  the 
expansion  of  (a  +  by. 

4.3.0 
Will  give  the  coefficient  of  the  fourth  term  of  the 

1  *  Z  *  o 

5.4.3 
expansion  of  (a  +  6)^?     Will       ^    o  give  the  fourth  co- 
efficient of  (a  +  by?    Write  the  fourth  term  of  (a  +  by. 

282 


XVIII.  §1211  BINOMIAL  THEOREM  -         283 

The  laws  just  brought  out  may  be  expressed  in   algebraic 
language : 

This  equation  expresses  what  is  known  as  the  Binomial 
Theorem. 

Letn  =  7; 

(a  +  by  =  a'  +  7a%  +  '~a'b^ 

.7-6.5   4,3  ,7.6.5.4   3,,  ^7.6.5.4.3   _ 

+ 172:3  ^  ^  +  r^^T^  ^  ^  + 1.2.3.4.5 «  ^ 

•''''''''^'Kb^  +  b^ 


'  1.2.3.4.5.6 
=  aJ+7  a'b  +21  a'b^-hS5  a'b'+  35  a^¥-{-  21  a%^  +  7a¥  +  b\ 

Example.  (a  -  2  62)5  =  ? 

In  the  formula  replace  n  by  5,  leave  a  unchanged,  and  replace  h  by 
-2  62.     Then 

(a -262)5  =  a5  4.|a4(_262)  ^^^a^(-2¥y 

=  a5  -  10  a*62  +  40  0^6^  -  80  a'Jfi  +  80  a¥  -  32  6^0. 

Exercises.     Write  expansions  of  the  following: 

1.  (2  a  -  by.  8.    .98^  =  (1  -  .02)4. 

2.  (r2  -  sy.  9.    l.OP  =  (1  +  .01)5. 

3.  (J  a  +  62)6,  jQ^     gge    _    (jQQ  _    -^)6, 

4.  /I    _2a2y.  11.    (3:^  +  ^)'. 

6.    (a-  +  ay.  13.    (..^  +  z;^)l 

/  ly  •       14.   (xV-1)'. 

^-    (^^         3^j-  15.    (2^3_i^i)6. 

16.  (a-5  -  2  Va)^ 


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